Test Objectives
  • Demonstrate the ability to find the exact value of a trigonometric expression
  • Demonstrate the ability to simplify a trigonometric expression
  • Demonstrate the ability to verify a trigonometric identity
Sum & Difference Identities for Sine & Tangent Practice Test:

#1:

Instructions: Find the exact value.

$$a)\hspace{.1em}\text{sin}\frac{5 π}{12}$$

$$b)\hspace{.1em}\text{tan}\hspace{.1em}255 °$$


#2:

Instructions: Find the exact value.

$$a)\hspace{.1em}\text{cos}\frac{47 π}{18}\hspace{.1em}\text{cos}\frac{10π}{9}+ \text{sin}\frac{47 π}{18}\hspace{.1em}\text{sin}\frac{10π}{9}$$

$$b)\hspace{.1em}\frac{\text{tan}\large{\frac{π}{18}}+ \text{tan}\large{\frac{5 π}{18}}}{1 - \text{tan}\large{\frac{π}{18}}\text{tan}\large{\frac{5 π}{18}}}$$


#3:

Instructions: Simplify.

$$a)\hspace{.1em}\text{sin}(-4v)\hspace{.1em}\text{cos}\hspace{.1em}5v + \text{cos}(-4v)\hspace{.1em}\text{sin}\hspace{.1em}5v$$

$$b)\hspace{.1em}\frac{\text{tan}\hspace{.1em}θ - \text{tan}\hspace{.1em}4θ}{1 + \text{tan}\hspace{.1em}θ \hspace{.1em}\text{tan}\hspace{.1em}4θ}$$


#4:

Instructions: Find sin(θ + β), tan(θ + β), and the quadrant of θ + β.

$$a)\hspace{.1em}\text{cos θ}=\frac{3}{5}$$ $$\text{sin β}=\frac{5}{13}$$ θ and β are in quadrant I

$$b)\hspace{.1em}\text{cos θ}=-\frac{8}{17}$$ $$\text{cos β}=-\frac{3}{5}$$ θ and β are in quadrant III


#5:

Instructions: Verify each identity.

$$a)\hspace{.1em}\frac{\text{tan}\hspace{.1em}θ + 1}{1 - \text{tan}\hspace{.1em}θ}=\text{tan}(θ - 135°)$$

$$b) \hspace{.1em}\frac{\text{cos}(θ - β)}{\text{cos θ}\hspace{.1em}\text{sin β}}=\text{tan}\hspace{.1em}θ + \text{cot}\hspace{.1em}β$$


Written Solutions:

#1:

Solutions:

$$a)\hspace{.1em}\frac{\sqrt{6}+ \sqrt{2}}{4}$$

$$b)\hspace{.1em}2 + \sqrt{3}$$


#2:

Solutions:

$$a)\hspace{.1em}0$$

$$b)\hspace{.1em}\sqrt{3}$$


#3:

Solutions:

$$a)\hspace{.1em}\text{sin}\hspace{.1em}v$$

$$b)\hspace{.1em}\text{tan}(-3θ)$$


#4:

Solutions:

$$a)\hspace{.1em}\text{sin}(θ + β)=\frac{63}{65}$$ $$\text{tan}(θ + β)=\frac{63}{16}$$ (θ + β) is in quadrant I

$$b)\hspace{.1em}\text{sin}(θ + β)=\frac{77}{85}$$ $$\text{tan}(θ + β)=-\frac{77}{36}$$ (θ + β) is in quadrant II


#5:

Solutions:

$$a)\hspace{.1em}\frac{\text{tan}\hspace{.1em}θ + 1}{1 - \text{tan}\hspace{.1em}θ}=\text{tan}(θ - 135°)$$ $$=\text{tan}(θ - 135°)$$ $$=\frac{\text{tan}\hspace{.1em}θ - (-1)}{1 + \text{tan θ}(-1)}$$ $$=\frac{\text{tan}\hspace{.1em}θ + 1}{1 - \text{tan}\hspace{.1em}θ}✓$$

$$b)\hspace{.1em}\frac{\text{cos}(θ - β)}{\text{cos θ}\hspace{.1em}\text{sin β}}=\text{tan}\hspace{.1em}θ + \text{cot}\hspace{.1em}β$$ $$\text{tan}\hspace{.1em}θ + \text{cot}\hspace{.1em}β=\frac{\text{cos}(θ - β)}{\text{cos θ}\hspace{.1em}\text{sin β}}$$ $$=\frac{\text{cos}\hspace{.1em}θ \hspace{.1em}\text{cos}β + \text{sin}\hspace{.1em}θ \hspace{.1em}\text{sin}\hspace{.1em}β}{\text{cos θ}\hspace{.1em}\text{sin β}}$$ $$=\frac{\text{cos}\hspace{.1em}θ \hspace{.1em}\text{cos}β}{\text{cos θ}\hspace{.1em}\text{sin β}}+ \frac{\text{sin}\hspace{.1em}θ \hspace{.1em}\text{sin}\hspace{.1em}β}{\text{cos θ}\hspace{.1em}\text{sin β}}$$ $$=\frac{\text{cos}β}{\text{sin β}}+ \frac{\text{sin}\hspace{.1em}θ}{\text{cos θ}}$$ $$=\text{cot β}+ \text{tan θ}$$ $$=\text{tan θ}+ \text{cot β}✓$$