Trigonometric Identities & Equations — Challenge

One identity comes straight from Pythagoras, and it unlocks nearly every trig equation.

What these problems cover

These questions use 10 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Problem 1

What does sin2x+cos2x equal?

  1. 0
  2. 1
  3. 2
  4. sin 2x
  5. tan x
Worked answer

1

Take a point on the unit circle: its coordinates are (cos x,sin x).

The circle has radius 1, so by Pythagoras cos2x+sin2x = 12.

So the expression equals 1 for every angle. This is the Pythagorean identity.

Problem 2

What does tan x equal in terms of sine and cosine?

  1. sin xcos x
  2. cos xsin x
  3. sin xcos x
  4. 1sin x
  5. sin x+cos x
Worked answer

sin xcos x

In a right triangle, sine is opposite over hypotenuse and cosine is adjacent over hypotenuse.

Dividing one by the other cancels the hypotenuse and leaves opposite over adjacent, which is the tangent.

So tan x = sin xcos x, undefined wherever cos x = 0.

Problem 3

If sinθ+cosθ=12, what is 2sinθcosθ?

  1. -34
  2. -14
  3. 14
  4. 34
  5. 1
Worked answer

-34

Square both sides: (sinθ+cosθ)2=sin2θ+2sinθcosθ+cos2θ. The Pythagorean identity turns sin2θ+cos2θ into 1, so the left side is 1+2sinθcosθ. That equals (12)2=14, so 2sinθcosθ=14-1=-34.

Problem 4

For sinθ ≠ 0, simplify 1-cos2θsinθ.

  1. sinθ
  2. cosθ
  3. tanθ
  4. 1
  5. cotθ
Worked answer

sinθ

The Pythagorean identity rearranges to 1-cos2θ=sin2θ. So the expression is sin2θsinθ. That cancels to sinθ.

Problem 5

If tanθ=34 and θ is acute, what is sinθ+cosθ?

  1. 1
  2. 75
  3. 57
  4. 125
  5. 712
Worked answer

75

A tangent of 34 means the opposite and adjacent sides are 3 and 4, so the hypotenuse is 5. Then sinθ=35 and cosθ=45. Adding: 35+45=75.

Problem 6

How many solutions does 2cos2θ-cosθ-1=0 have for 0° ≤ θ < 360°?

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
Worked answer

3

Let c=cosθ. Then 2c2-c-1=0 factors as (2c+1)(c-1)=0. cosθ=-12 gives two angles, 120° and 240°. cosθ=1 gives one more, θ=0°. That is 3 altogether.

Problem 7

Simplify sinθ · cotθ.

  1. sinθ
  2. cosθ
  3. tanθ
  4. 1
  5. secθ
Worked answer

cosθ

Write cotθ as cosθsinθ. Then sinθ · cosθsinθ. The sines cancel, leaving cosθ.

Problem 8

If secθ=53 and θ is in quadrant IV, what is tanθ?

  1. 43
  2. -43
  3. 34
  4. -34
  5. -54
Worked answer

-43

secθ=53 means cosθ=35. From the identity, sin2θ=1-925=1625, so sinθ=±45. Quadrant IV has y negative, so sinθ=-45. Then tanθ=-4/53/5=-43.

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