Unit 9 Review: Trig Equations & Identities

 $1-\cos^2x=?$  

•  $\cot^2x$  

•  $\tan^2x$  

•  $\csc^2x$  

•  $\sin^2x$  

Answer explanation

Remember the Pythagorean Identity:  $1=\sin^2x+\cos^2x$ 
In this question, all they did was subtract  $\cos^2x$   from both sides. Can you see that?

Which of the following is a solution of the equation

 $\sqrt{3}\csc x=2$  

•  $\frac{\pi}{6}$  

•  $\frac{\pi}{3}$  

•  $\frac{\pi}{4}$  

•  $\pi$  

Answer explanation

Step 1: Isolate Trig Function (Divide both sides by square root of 3)
 $\frac{\sqrt{3}\csc x}{\sqrt{3}}=\ \frac{2}{\sqrt{3}}$  
Then find reciprocal identity of csc which is inverse of sin (y value) to evaluate on Unit Circle:
 $\sin=\frac{\sqrt{3}}{2}$  =  $\frac{\pi}{3},\ \frac{2\pi}{3}$  

Tags

CCSS.HSF.TF.B.7

Simplify

• -1
• sin θ
• csc θ
• 1

Simplify (Hint: you will need to FOIL first)

• sinθ

• cot²θ

• tan²θ

• cos²θ

Solve
cosθ = - √(3)/2  
on θ∈[0, 2π)

• θ = π /3, 2π /3
• θ = 2π /3, 4π /3
• θ = π /6, 5π /6
• θ = 5π /6, 7π /6

Tags

CCSS.HSF.TF.B.7

2 sinθ+3=2

• π/6,2π/3
• 7π/6 
• 7π/6, 11π/6

Tags

CCSS.HSF.TF.B.7

Solve on the domain [0, 2π)
4sin²x = 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

Tags

CCSS.HSF.TF.B.7

Solve on the Interval [0,2π)
     tan(x) + 1 = 0     

• x = π/4, x = 5π/4
• x = π/4, x = 3π/4
• x = π/2
• x = 3π/4, x = 7π/4

Tags

CCSS.HSF.TF.B.7

Multiply and simplify.
(1 - cos x)(1 + cos x)

• sin² x
• cos² x
• tan² x
• csc² x

cos²x + sin²x = 3

• no solution

• 1

• tan²x

• π²

Answer explanation

Remember: cos²x + sin²x = 1

Tags

CCSS.HSF.TF.B.7