Final Exam Review: Unit 7 (Trig Identities and Equations)

• ​Simplify and verify trig identities

• Use standard algebraic techniques to solve trig equations

  • Collecting like terms

  • GCF/factoring

  • Using square roots

  • Using Pythaogrean identities

​Unit 7: Trig Identities and Equations

​Link to Notes

Multiple Choice

Complete the identity...

$$1+\tan^2x=$$  

• $$\csc^2x$$  

• $$\cot^2x$$  

• $$\sin^2x$$  

• $$\sec^2x$$  

Multiple Choice

Complete the identity...

$$1-\sin^2x\ =$$  

• $$\sin^2x$$  

• $$\cos^2x$$  

• $$\sec^2x$$  

• $$\csc^2x$$  

Fill in the Blanks

Multiple Choice

Simplify. $$\sec x\cdot\cot x\cdot\sin x$$  

• -1

• secx

• cosx

• 1

Match

Match the following

• $$\frac{1}{\csc\theta}$$

• $$\frac{1}{\cos\theta}$$

• $$\frac{\sin\theta}{\cos\theta}$$

• $$\frac{1}{\sin\theta}$$

• $$\frac{\cos\theta}{\sin\theta}$$

sinθ

secθ

tanθ

cscθ

cotθ

• sinθ

• secθ

• tanθ

• cscθ

• cotθ

Multiple Choice

Please select the correct solution

• csc x

• sec x

• cot x

• tan x

Multiple Choice

What is csc(x) equivalent to?

• $$\frac{1}{\sin x}$$

• $$\frac{1}{\tan x}$$

• sin(x)

• $$\frac{1}{\cos x}$$

Multiple Choice

Simplify:  $$\left(\cos\ θ\right)\left(\sec\ θ-\cosθ\right)$$  

•   $$\sin^2θ$$  

• $$\cos^2θ$$  

• $$\csc^2\theta$$  

• $$\sec^2\theta$$  

Multiple Choice

Simplify:  $$(\secθ-1)(\secθ+1)$$  

• sin ² θ

• cos ² θ

• tan ² θ

• sec ² θ

Multiple Choice

Simplify:   $$\cos\theta\tan\theta$$  

• sin θ

• cos θ

• tan θ

• csc θ

Multiple Choice

To verify, which side seems the most complex.

• Left

• Right

Multiple Choice

Solve on the domain [0, 2π)
$$4+4\tan\theta=0$$  

• $$\theta=\pi$$  

• $$\theta=\frac{7\pi}{4}$$  

• $$\theta=\frac{3\pi}{4},\frac{7\pi}{4}$$  

• $$\theta=0,\frac{3\pi}{4}$$  

Multiple Choice

Solve on the domain [0, 2π)
$$2=-4-3\csc\theta$$  

• $$\theta=\frac{2\pi}{3},\frac{7\pi}{6}$$  

• $$\theta=\frac{\pi}{3}$$  

• $$\theta=\frac{7\pi}{6},\frac{11\pi}{6}$$  

• $$\theta=\frac{2\pi}{3},\frac{7\pi}{6},\frac{11\pi}{6}$$  

Multiple Choice

$$Solve\ \ 2\sin^2x\ +\ \sin x\ -\ 1\ =\ 0\ \ for\ 0\ \le\ x\ <\ 2\pi$$  

• $$\frac{\pi}{6},\ \frac{\pi}{2},\ \frac{5\pi}{6}$$  

• $$\frac{\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}$$  

• $$\frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{3\pi}{2}$$  

• $$\frac{7\pi}{6},\ \frac{3\pi}{2},\ \frac{11\pi}{6}$$  

Multiple Choice

Solve sinθ = ½ on θ∈[0, 2π)

• θ = π / 6,  7π / 6
• θ = π / 6,  5π / 6
• θ = 5π / 6,  7π / 6
• θ = 7π / 6,  11π / 6

Multiple Choice

Solve on the domain [0, 2π)
$$\cosθ=1$$

• θ = 0

• θ = π/2

• θ = π

• θ = 3π/2

Multiple Choice

Which of these is equivalent to $$2\cos^2x−3\cos x=0$$ ?

• -cos²x = 0

• cosx(2cosx + 3) = 0

• cosx(2cosx − 3) = 0

• cos x = ⅔

Multiple Choice

Why doesn't $$2\cos x−3=0$$ have solutions?

• cos x is never bigger than one

• cos x is never equal to a fraction

• Actually, this equation does have a solution, x = π

• This equation will have a solution tomorrow.

Multiple Choice

To solve this equation, $$\cos^2x+\sin x=1$$ replace $$\cos^2x$$ with

• 1/(sec²x)

• sin²x − 1

• 1 − sin²x

• 1 + tan²x

Multiple Choice

$$\csc^2x=2$$ is equivalent to $$\sin^2x=\frac{1}{2}$$

• True

• False

Multiple Choice

$$\cos x+\cos^2x=0$$  

What method would you use to solve the equation?

• greatest common factor

• cancel cosines

• combine like terms

• square roots

Multiple Choice

Solve on the domain [0, 2π)
$$4\sin^2x=3$$

• π/6, 11π/6
• π/3, 2π/3
• π/6, 5π/6, 7π/6, 11π/6
• π/3, 2π/3, 4π/3, 5π/3

Multiple Choice

Solve on the domain [0, 2π)

$$\sqrt{2}\cos x-1=0$$  

• π/4

• π/4 and  7π/4

• 7π/4

• π/4, 3π/4