Trigonometric identities and equations

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1. The Pythagorean identity and simplifying

Warm-up

Three quick questions from earlier lessons.

Discover

The point (cos⁡x,sin⁡x)(\cos x, \sin x) is on the unit circle, whose equation is X2+Y2=1X^2 + Y^2 = 1. What does that tell you?

  1. cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1
  2. cos⁡x+sin⁡x=1\cos x + \sin x = 1

Key idea

An identity is an equation that is true for every value of the variable.

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, so cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x and sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x. Also tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x}. Use them to find one value from another (the quadrant gives the sign) and to simplify.

Worked example

cos⁡θ=−35\cos\theta = -\dfrac{3}{5} and θ\theta is in quadrant III. Find sin⁡θ\sin\theta.

sin⁡2θ=1−925=1625\sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}
The Pythagorean identity.
sin⁡θ=±45\sin\theta = \pm\dfrac{4}{5}
Two square roots are possible.
sin⁡θ=−45\sin\theta = -\dfrac{4}{5}
Sine is negative in quadrant III.

Your turn, guided

Your turn: sin⁡θ=513\sin\theta = \dfrac{5}{13} and θ\theta is in quadrant I.

cos⁡2θ=144169\cos^2\theta = \dfrac{144}{169}
1 − 25/169.
cos⁡θ=1213\cos\theta = \dfrac{12}{13}
Positive in quadrant I.

Practice ★

  1. sin⁡θ=15\sin\theta = \frac{1}{5} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  2. sin⁡θ=38\sin\theta = \frac{3}{8} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  3. sin⁡θ=16\sin\theta = \frac{1}{6} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  4. sin⁡θ=13\sin\theta = \frac{1}{3} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.

Watch out!

Practice ★★

  1. Which expression is equal to 1−cos⁡2xsin⁡x\dfrac{1 - \cos^2 x}{\sin x} (where sin⁡x≠0\sin x \ne 0)?
    1. sin⁡x\sin x
    2. cos⁡x\cos x
    3. 1−cos⁡x1 - \cos x
  1. Which equation is true for every xx?
    1. sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1
    2. sin⁡x+cos⁡x=1\sin x + \cos x = 1
    3. sin⁡2x−cos⁡2x=1\sin^2 x - \cos^2 x = 1
    4. tan⁡x=cos⁡xsin⁡x\tan x = \dfrac{\cos x}{\sin x}

Exit ticket

Three questions to check you have got it.

  1. sin⁡θ=15\sin\theta = \frac{1}{5} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  2. sin⁡θ=38\sin\theta = \frac{3}{8} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  3. sin⁡θ=16\sin\theta = \frac{1}{6} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.

Summary card

The Pythagorean identity

  • sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 for every xx.
  • From one value: square, subtract from 1, take the root, choose the sign by quadrant.

Watch out: the identity needs the squares.

2. Solve sin x = c, cos x = c and tan x = c

Warm-up

Three quick questions from earlier lessons.

Discover

How many angles xx with 0≤x<2π0 \le x < 2\pi have sin⁡x=12\sin x = \dfrac{1}{2}?

  1. Two
  2. One

Key idea

On [0,2π)[0, 2\pi), an equation like sin⁡x=c\sin x = c usually has two solutions.

Find the reference angle α\alpha (from the absolute value of cc). Then use the quadrants where the function has the sign of cc: sine in I, II (positive) or III, IV (negative); cosine in I, IV or II, III; tangent in I, III or II, IV. The solutions are α\alpha, π−α\pi - \alpha, π+α\pi + \alpha or 2π−α2\pi - \alpha, depending on the quadrant.

Worked example

Solve cos⁡x=−22\cos x = -\dfrac{\sqrt{2}}{2} for 0≤x<2π0 \le x < 2\pi.

Reference angle: π4\dfrac{\pi}{4}
cos(π/4) = √2/2.
x=π−π4x = \pi - \dfrac{\pi}{4} or x=π+π4x = \pi + \dfrac{\pi}{4}
Cosine is negative in quadrants II and III.
x=3π4x = \dfrac{3\pi}{4} or x=5π4x = \dfrac{5\pi}{4}
Simplify.

Your turn, guided

Your turn: sin⁡x=−12\sin x = -\dfrac{1}{2}.

Reference angle π6\dfrac{\pi}{6}: x=π+π6x = \pi + \dfrac{\pi}{6} or x=2π−π6x = 2\pi - \dfrac{\pi}{6}
sin(π/6) = 1/2; sine is negative in quadrants III and IV.
x=7π6x = \dfrac{7\pi}{6} or x=11π6x = \dfrac{11\pi}{6}
Type the quadrant III solution.

Practice ★

  1. Solve sin⁡x=12\sin x = \frac{1}{2} for 0≤x<2π0 \le x < 2\pi.
  2. Solve sin⁡x=22\sin x = \frac{\sqrt{2}}{2} for 0≤x<2π0 \le x < 2\pi.
  3. Solve sin⁡x=32\sin x = \frac{\sqrt{3}}{2} for 0≤x<2π0 \le x < 2\pi.

Watch out!

Practice ★★

  1. Solve cos⁡x=−22\cos x = - \frac{\sqrt{2}}{2} for 0≤x<2π0 \le x < 2\pi.
  2. Solve cos⁡x=−12\cos x = - \frac{1}{2} for 0≤x<2π0 \le x < 2\pi.
  1. Solve tan⁡x=3\tan x = \sqrt{3} for 0≤x<2π0 \le x < 2\pi.
  2. Solve tan⁡x=33\tan x = \frac{\sqrt{3}}{3} for 0≤x<2π0 \le x < 2\pi.

Exit ticket

Three questions to check you have got it.

  1. Solve cos⁡x=−22\cos x = - \frac{\sqrt{2}}{2} for 0≤x<2π0 \le x < 2\pi.
  2. Solve cos⁡x=−12\cos x = - \frac{1}{2} for 0≤x<2π0 \le x < 2\pi.
  3. Solve cos⁡x=−32\cos x = - \frac{\sqrt{3}}{2} for 0≤x<2π0 \le x < 2\pi.

Summary card

Solving sin x = c, cos x = c, tan x = c

  1. Reference angle from ∣c∣|c|.
  2. The two quadrants with the right sign.
  3. Write both solutions in [0,2π)[0, 2\pi).

Watch out: there are usually two solutions, not one.

3. Solve trigonometric equations by factoring

Warm-up

Three quick questions from earlier lessons.

Discover

To solve sin⁡xcos⁡x=sin⁡x\sin x \cos x = \sin x, Leo divided both sides by sin⁡x\sin x. Is that safe?

  1. No: it loses the solutions where sin⁡x=0\sin x = 0
  2. Yes, it is always allowed

Key idea

Treat sin⁡x\sin x or cos⁡x\cos x like an unknown, then factor.

Bring every term to one side and factor (never divide by sin⁡x\sin x or cos⁡x\cos x). A product is 0 when a factor is 0. For a quadratic in sin⁡x\sin x, let u=sin⁡xu = \sin x, solve for uu, then solve each sin⁡x=u\sin x = u (no solution if ∣u∣>1|u| > 1).

Worked example

Solve 2sin⁡2x−sin⁡x=02\sin^2 x - \sin x = 0 for 0≤x<2π0 \le x < 2\pi.

sin⁡x (2sin⁡x−1)=0\sin x\,(2\sin x - 1) = 0
Common factor sin x.
sin⁡x=0\sin x = 0: x=0x = 0 or x=πx = \pi
First factor.
x=π6x = \dfrac{\pi}{6} or x=5π6x = \dfrac{5\pi}{6}
Second factor: sin x = 1/2.

Your turn, guided

Your turn: cos⁡2x−1=0\cos^2 x - 1 = 0.

cos⁡x=1\cos x = 1 or cos⁡x=−1\cos x = -1
cos²x = 1.
x=0x = 0 or x=πx = \pi
cos x = −1: type the angle.

Practice ★

  1. Solve sin⁡x(sin⁡x−32)=0\sin x\left(\sin x - \frac{\sqrt{3}}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.
  2. Solve sin⁡x(sin⁡x−22)=0\sin x\left(\sin x - \frac{\sqrt{2}}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.
  3. Solve sin⁡x(sin⁡x−12)=0\sin x\left(\sin x - \frac{1}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.

Watch out!

Practice ★★

  1. Solve 3cos⁡(x)+cos⁡(2x)+2=03 \cos{\left(x \right)} + \cos{\left(2 x \right)} + 2 = 0 for 0≤x<2π0 \le x < 2\pi.
  2. Solve −3cos⁡(x)+cos⁡(2x)+2=0- 3 \cos{\left(x \right)} + \cos{\left(2 x \right)} + 2 = 0 for 0≤x<2π0 \le x < 2\pi.
  1. Put the steps for solving 2sin⁡2x+sin⁡x−1=02\sin^2 x + \sin x - 1 = 0 in order.
    • Let u=sin⁡xu = \sin x: 2u2+u−1=02u^2 + u - 1 = 0
    • Factor: (2u−1)(u+1)=0(2u - 1)(u + 1) = 0
    • sin⁡x=12\sin x = \dfrac{1}{2} or sin⁡x=−1\sin x = -1
    • List every xx in [0,2π)[0, 2\pi): π6,5π6,3π2\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}

Exit ticket

Three questions to check you have got it.

  1. Solve sin⁡x(sin⁡x−32)=0\sin x\left(\sin x - \frac{\sqrt{3}}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.
  2. Solve sin⁡x(sin⁡x−22)=0\sin x\left(\sin x - \frac{\sqrt{2}}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.
  3. Solve sin⁡x(sin⁡x−12)=0\sin x\left(\sin x - \frac{1}{2}\right) = 0 for 0≤x<2π0 \le x < 2\pi.

Summary card

Solving by factoring

  1. One side 0, then factor (or set u=sin⁡xu = \sin x).
  2. Each factor = 0.
  3. Solve each sin⁡x=u\sin x = u in [0,2π)[0, 2\pi).

Watch out: never divide by sin⁡x\sin x or cos⁡x\cos x.

4. Equations with a multiple angle

Warm-up

Three quick questions from earlier lessons.

Discover

sin⁡2x\sin 2x repeats twice as fast as sin⁡x\sin x. On [0,2π)[0, 2\pi), how many solutions does sin⁡2x=12\sin 2x = \dfrac{1}{2} have?

  1. Four
  2. Two

Key idea

Solve for the whole angle first, on the stretched interval.

For sin⁡(nx)=c\sin(nx) = c with 0≤x<2π0 \le x < 2\pi: let θ=nx\theta = nx, which runs over [0,2nπ)[0, 2n\pi). Find every θ\theta there (add 2π2\pi again and again), then divide each by nn.

Worked example

Solve cos⁡2x=0\cos 2x = 0 for 0≤x<2π0 \le x < 2\pi.

cos⁡θ=0\cos\theta = 0: θ=π2,3π2,5π2,7π2\theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2}, \dfrac{7\pi}{2}
θ = 2x runs over [0, 4π).
x=π4,3π4,5π4,7π4x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}
Divide each by 2.

Your turn, guided

Your turn: sin⁡2x=1\sin 2x = 1.

θ=π2\theta = \dfrac{\pi}{2} or 5π2\dfrac{5\pi}{2}
θ = 2x in [0, 4π): sin θ = 1 at π/2 and π/2 + 2π.
x=π4x = \dfrac{\pi}{4} or x=5π4x = \dfrac{5\pi}{4}
Divide 5π/2 by 2.

Practice ★

  1. How many solutions does sin⁡(5x)=12\sin(5x) = \dfrac{1}{2} have for 0≤x<2π0 \le x < 2\pi?
  2. How many solutions does sin⁡(4x)=12\sin(4x) = \dfrac{1}{2} have for 0≤x<2π0 \le x < 2\pi?
  3. How many solutions does sin⁡(2x)=12\sin(2x) = \dfrac{1}{2} have for 0≤x<2π0 \le x < 2\pi?

Watch out!

Practice ★★

  1. Solve sin⁡2x=12\sin 2x = \frac{1}{2} for 0≤x<2π0 \le x < 2\pi.
  2. Solve sin⁡2x=32\sin 2x = \frac{\sqrt{3}}{2} for 0≤x<2π0 \le x < 2\pi.
  1. Solve cos⁡3x=−1\cos 3x = -1 for 0≤x<2π0 \le x < 2\pi.

Exit ticket

Three questions to check you have got it.

  1. How many solutions does sin⁡(5x)=12\sin(5x) = \dfrac{1}{2} have for 0≤x<2π0 \le x < 2\pi?
  2. How many solutions does sin⁡(4x)=12\sin(4x) = \dfrac{1}{2} have for 0≤x<2π0 \le x < 2\pi?
  3. How many solutions does sin⁡(2x)=12\sin(2x) = \dfrac{1}{2} have for 0≤x<2π0 \le x < 2\pi?

Summary card

Multiple angles

  1. Let θ=nx\theta = nx: it runs over [0,2nπ)[0, 2n\pi).
  2. List every θ\theta there.
  3. Divide each by nn.

Watch out: sin⁡2x\sin 2x has twice as many solutions as sin⁡x\sin x.

5. Sum and difference identities

Warm-up

Three quick questions from earlier lessons.

Discover

Is cos⁡(30∘+60∘)\cos(30^\circ + 60^\circ) equal to cos⁡30∘+cos⁡60∘\cos 30^\circ + \cos 60^\circ?

  1. No
  2. Yes

Key idea

The sum and difference identities give exact values for angles like 75∘75^\circ or 15∘15^\circ.

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B. cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B (the sign flips for cosine).

Worked example

Find the exact value of sin⁡75∘\sin 75^\circ.

sin⁡75∘=sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ
75° = 45° + 30°.
=22⋅32+22⋅12= \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}
Special values.
=6+24= \dfrac{\sqrt{6} + \sqrt{2}}{4}
Common denominator.

Your turn, guided

Your turn: cos⁡15∘=cos⁡(45∘−30∘)\cos 15^\circ = \cos(45^\circ - 30^\circ).

=cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘= \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ
For a difference, cosine takes a plus.
=6+24= \dfrac{\sqrt{6} + \sqrt{2}}{4}
√6/4 + √2/4.

Practice ★

  1. Use a sum identity to find the exact value of cos⁡(45∘+30∘)\cos(45^\circ + 30^\circ).
  2. Use a sum identity to find the exact value of cos⁡(60∘+30∘)\cos(60^\circ + 30^\circ).
  3. Use a sum identity to find the exact value of cos⁡(60∘+45∘)\cos(60^\circ + 45^\circ).

Watch out!

Practice ★★

  1. Which expression is equal to cos⁡(A−B)\cos(A - B)?
    1. cos⁡Acos⁡B+sin⁡Asin⁡B\cos A\cos B + \sin A\sin B
    2. cos⁡Acos⁡B−sin⁡Asin⁡B\cos A\cos B - \sin A\sin B
    3. cos⁡A−cos⁡B\cos A - \cos B
  1. AA and BB are in quadrant I, sin⁡A=35\sin A = \frac{3}{5} and cos⁡B=513\cos B = \dfrac{5}{13}. Find the exact value of sin⁡(A+B)\sin(A + B).
  2. AA and BB are in quadrant I, sin⁡A=725\sin A = \frac{7}{25} and cos⁡B=513\cos B = \dfrac{5}{13}. Find the exact value of sin⁡(A+B)\sin(A + B).

Exit ticket

Three questions to check you have got it.

  1. Use a sum identity to find the exact value of cos⁡(45∘+30∘)\cos(45^\circ + 30^\circ).
  2. Use a sum identity to find the exact value of cos⁡(60∘+30∘)\cos(60^\circ + 30^\circ).
  3. Use a sum identity to find the exact value of cos⁡(60∘+45∘)\cos(60^\circ + 45^\circ).

Summary card

Sum and difference identities

  • sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B.
  • cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B.

Watch out: cos⁡(A+B)≠cos⁡A+cos⁡B\cos(A + B) \ne \cos A + \cos B; cosine flips the sign.

6. Double-angle identities

Warm-up

Three quick questions from earlier lessons.

Discover

Is sin⁡2x\sin 2x the same as 2sin⁡x2\sin x? Try x=π2x = \dfrac{\pi}{2}.

  1. No
  2. Yes

Key idea

Put A=B=xA = B = x in the sum identities.

sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x. cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x.

Worked example

sin⁡θ=35\sin\theta = \dfrac{3}{5}, θ\theta in quadrant I. Find sin⁡2θ\sin 2\theta and cos⁡2θ\cos 2\theta.

cos⁡θ=45\cos\theta = \dfrac{4}{5}
Pythagorean identity; positive in quadrant I.
sin⁡2θ=2⋅35⋅45=2425\sin 2\theta = 2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5} = \dfrac{24}{25}
2 sin θ cos θ.
cos⁡2θ=1−2⋅925=725\cos 2\theta = 1 - 2\cdot\dfrac{9}{25} = \dfrac{7}{25}
1 − 2 sin²θ.

Your turn, guided

Your turn: cos⁡θ=13\cos\theta = \dfrac{1}{3}. Find cos⁡2θ\cos 2\theta.

cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1
The form with cosine.
=−79= -\dfrac{7}{9}
2 · 1/9 − 1.

Practice ★

  1. cos⁡θ=37\cos\theta = \frac{3}{7}. Find the exact value of cos⁡2θ\cos 2\theta.
  2. cos⁡θ=17\cos\theta = \frac{1}{7}. Find the exact value of cos⁡2θ\cos 2\theta.
  3. cos⁡θ=25\cos\theta = \frac{2}{5}. Find the exact value of cos⁡2θ\cos 2\theta.

Watch out!

Practice ★★

  1. θ\theta is in quadrant I and sin⁡θ=1517\sin\theta = \frac{15}{17}. Find the exact value of sin⁡2θ\sin 2\theta.
  2. θ\theta is in quadrant I and sin⁡θ=35\sin\theta = \frac{3}{5}. Find the exact value of sin⁡2θ\sin 2\theta.
  1. Which expressions are equal to cos⁡2x\cos 2x? Tick all of them.
    1. cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x
    2. 2cos⁡2x−12\cos^2 x - 1
    3. 1−2sin⁡2x1 - 2\sin^2 x
    4. 2cos⁡x2\cos x
    5. cos⁡2x+sin⁡2x\cos^2 x + \sin^2 x

Exit ticket

Three questions to check you have got it.

  1. cos⁡θ=37\cos\theta = \frac{3}{7}. Find the exact value of cos⁡2θ\cos 2\theta.
  2. cos⁡θ=17\cos\theta = \frac{1}{7}. Find the exact value of cos⁡2θ\cos 2\theta.
  3. cos⁡θ=25\cos\theta = \frac{2}{5}. Find the exact value of cos⁡2θ\cos 2\theta.

Summary card

Double-angle identities

  • sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x.
  • cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x.

Watch out: sin⁡2x≠2sin⁡x\sin 2x \ne 2\sin x (a sine is never above 1, but 2sin⁡x2\sin x can be).

7. Solve equations with identities

Warm-up

Three quick questions from earlier lessons.

Discover

To solve sin⁡2x=sin⁡x\sin 2x = \sin x, what is a good first step?

  1. Replace sin⁡2x\sin 2x by 2sin⁡xcos⁡x2\sin x\cos x
  2. Write sin⁡2x=2sin⁡x\sin 2x = 2\sin x
  3. Divide both sides by sin⁡x\sin x

Key idea

Identities turn an equation with different angles into one you can factor.

Rewrite with one angle (sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x; cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1 or 1−2sin⁡2x1 - 2\sin^2 x, whichever matches the other terms). Then bring everything to one side, factor, and solve each factor on [0,2π)[0, 2\pi).

Worked example

Solve sin⁡2x=sin⁡x\sin 2x = \sin x for 0≤x<2π0 \le x < 2\pi.

2sin⁡xcos⁡x−sin⁡x=02\sin x\cos x - \sin x = 0
Double-angle identity.
sin⁡x (2cos⁡x−1)=0\sin x\,(2\cos x - 1) = 0
Common factor, never divide.
x=0x = 0 or x=πx = \pi
sin x = 0.
x=π3x = \dfrac{\pi}{3} or x=5π3x = \dfrac{5\pi}{3}
cos x = 1/2.

Your turn, guided

Your turn: cos⁡2x+cos⁡x=0\cos 2x + \cos x = 0, using cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1.

2cos⁡2x+cos⁡x−1=02\cos^2 x + \cos x - 1 = 0
A quadratic in cos x.
cos⁡x=12\cos x = \dfrac{1}{2} or cos⁡x=−1\cos x = -1
(2cos x − 1)(cos x + 1) = 0: type the positive value of cos x.

Practice ★

  1. Solve sin⁡2x=sin⁡(x)\sin 2x = \sin{\left(x \right)} for 0≤x<2π0 \le x < 2\pi.
  2. Solve sin⁡2x=−sin⁡(x)\sin 2x = - \sin{\left(x \right)} for 0≤x<2π0 \le x < 2\pi.

Watch out!

Practice ★★

  1. Solve cos⁡2x=cos⁡x\cos 2x = \cos x for 0≤x<2π0 \le x < 2\pi.

Exit ticket

Three questions to check you have got it.

  1. Solve sin⁡2x=sin⁡(x)\sin 2x = \sin{\left(x \right)} for 0≤x<2π0 \le x < 2\pi.
  2. Solve sin⁡2x=−sin⁡(x)\sin 2x = - \sin{\left(x \right)} for 0≤x<2π0 \le x < 2\pi.

Summary card

Equations with identities

  1. Rewrite with one angle (double-angle identities).
  2. One side 0, factor (or u=sin⁡xu = \sin x, u=cos⁡xu = \cos x).
  3. Solve each factor in [0,2π)[0, 2\pi).

Watch out: factor, never divide by sin⁡x\sin x or cos⁡x\cos x.

Answers: Trigonometric identities and equations

1. The Pythagorean identity and simplifying

A

  1. −265- \frac{2 \sqrt{6}}{5}
  2. −558- \frac{\sqrt{55}}{8}
  3. −356- \frac{\sqrt{35}}{6}
  4. −223- \frac{2 \sqrt{2}}{3}

B

  1. sin⁡x\sin x

C

  1. sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

D

  1. −265- \frac{2 \sqrt{6}}{5}
  2. −558- \frac{\sqrt{55}}{8}
  3. −356- \frac{\sqrt{35}}{6}

2. Solve sin x = c, cos x = c and tan x = c

A

  1. π6, 5π6\frac{\pi}{6},\ \frac{5 \pi}{6}
  2. π4, 3π4\frac{\pi}{4},\ \frac{3 \pi}{4}
  3. π3, 2π3\frac{\pi}{3},\ \frac{2 \pi}{3}

B

  1. 3π4, 5π4\frac{3 \pi}{4},\ \frac{5 \pi}{4}
  2. 2π3, 4π3\frac{2 \pi}{3},\ \frac{4 \pi}{3}

C

  1. π3, 4π3\frac{\pi}{3},\ \frac{4 \pi}{3}
  2. π6, 7π6\frac{\pi}{6},\ \frac{7 \pi}{6}

D

  1. 3π4, 5π4\frac{3 \pi}{4},\ \frac{5 \pi}{4}
  2. 2π3, 4π3\frac{2 \pi}{3},\ \frac{4 \pi}{3}
  3. 5π6, 7π6\frac{5 \pi}{6},\ \frac{7 \pi}{6}

3. Solve trigonometric equations by factoring

A

  1. 0, π, π3, 2π30,\ \pi,\ \frac{\pi}{3},\ \frac{2 \pi}{3}
  2. 0, π, π4, 3π40,\ \pi,\ \frac{\pi}{4},\ \frac{3 \pi}{4}
  3. 0, π, π6, 5π60,\ \pi,\ \frac{\pi}{6},\ \frac{5 \pi}{6}

B

  1. π, 2π3, 4π3\pi,\ \frac{2 \pi}{3},\ \frac{4 \pi}{3}
  2. 0, π3, 5π30,\ \frac{\pi}{3},\ \frac{5 \pi}{3}

C

  1. 1. Let u=sin⁡xu = \sin x: 2u2+u−1=02u^2 + u - 1 = 0 2. Factor: (2u−1)(u+1)=0(2u - 1)(u + 1) = 0 3. sin⁡x=12\sin x = \dfrac{1}{2} or sin⁡x=−1\sin x = -1 4. List every xx in [0,2π)[0, 2\pi): π6,5π6,3π2\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}

D

  1. 0, π, π3, 2π30,\ \pi,\ \frac{\pi}{3},\ \frac{2 \pi}{3}
  2. 0, π, π4, 3π40,\ \pi,\ \frac{\pi}{4},\ \frac{3 \pi}{4}
  3. 0, π, π6, 5π60,\ \pi,\ \frac{\pi}{6},\ \frac{5 \pi}{6}

4. Equations with a multiple angle

A

  1. 1010
  2. 88
  3. 44

B

  1. π12, 5π12, 13π12, 17π12\frac{\pi}{12},\ \frac{5 \pi}{12},\ \frac{13 \pi}{12},\ \frac{17 \pi}{12}
  2. π6, π3, 7π6, 4π3\frac{\pi}{6},\ \frac{\pi}{3},\ \frac{7 \pi}{6},\ \frac{4 \pi}{3}

C

  1. π3, π, 5π3\frac{\pi}{3},\ \pi,\ \frac{5 \pi}{3}

D

  1. 1010
  2. 88
  3. 44

5. Sum and difference identities

A

  1. −24+64- \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}
  2. 00
  3. −64+24- \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}

B

  1. cos⁡Acos⁡B+sin⁡Asin⁡B\cos A\cos B + \sin A\sin B

C

  1. 6365\frac{63}{65}
  2. 323325\frac{323}{325}

D

  1. −24+64- \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}
  2. 00
  3. −64+24- \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}

6. Double-angle identities

A

  1. −3149- \frac{31}{49}
  2. −4749- \frac{47}{49}
  3. −1725- \frac{17}{25}

B

  1. 240289\frac{240}{289}
  2. 2425\frac{24}{25}

C

  1. cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x · 2cos⁡2x−12\cos^2 x - 1 · 1−2sin⁡2x1 - 2\sin^2 x

D

  1. −3149- \frac{31}{49}
  2. −4749- \frac{47}{49}
  3. −1725- \frac{17}{25}

7. Solve equations with identities

A

  1. 0, π, π3, 5π30,\ \pi,\ \frac{\pi}{3},\ \frac{5 \pi}{3}
  2. 0, π, 2π3, 4π30,\ \pi,\ \frac{2 \pi}{3},\ \frac{4 \pi}{3}

B

  1. 0, 2π3, 4π30,\ \frac{2 \pi}{3},\ \frac{4 \pi}{3}

C

  1. 0, π, π3, 5π30,\ \pi,\ \frac{\pi}{3},\ \frac{5 \pi}{3}
  2. 0, π, 2π3, 4π30,\ \pi,\ \frac{2 \pi}{3},\ \frac{4 \pi}{3}