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Hi, can you help me with problem D? I’m in high school calculus 1. Thank you!

Hi, can you help me with problem D? I’m in high school calculus 1. Thank you!-example-1
User Aexyn
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1 Answer

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Applying the next trigonometric identity:


\sin ^2\alpha+\cos ^2\alpha=1

to angle θ, and solving for cos(θ), we get:


\begin{gathered} \sin ^2\theta+\cos ^2\theta=1 \\ \text{ Substituting with }\sin ^{}\theta=1/3\colon \\ ((1)/(3))^2+\cos ^2\theta=1 \\ (1)/(9)+\cos ^2\theta=1 \\ \cos ^2\theta=1-(1)/(9) \\ \cos ^2\theta=(8)/(9) \\ \cos \theta=\sqrt[]{(8)/(9)} \\ \cos \theta=\frac{\sqrt[]{8}}{3} \\ \cos \theta=\frac{2\sqrt[]{2}}{3} \end{gathered}

The relation between the cosine and the secant of an angle is:


\begin{gathered} \sec \varphi=(1)/(\cos\varphi) \\ \text{ Substituting with }\sec \varphi=17/8\colon \\ (17)/(8)=(1)/(\cos\varphi) \\ ((17)/(8))^(-1)=((1)/(\cos\varphi))^(-1) \\ (8)/(17)=\cos \varphi \end{gathered}

Applying the before mentioned trigonometric identity to angle φ, we get:


\begin{gathered} \sin ^2\varphi+\cos ^2\varphi=1 \\ \text{ Substituting with cos}\varphi\text{ = 8/17}\colon \\ \sin ^2\varphi+((8)/(17))^2=1 \\ \sin ^2\varphi+(64)/(289)^{}=1 \\ \sin ^2\varphi=1-(64)/(289) \\ \sin ^2\varphi=(225)/(289) \\ \sin \varphi=\sqrt[]{(225)/(289)} \\ \sin \varphi=(15)/(17) \end{gathered}

Difference formula of cosine


\cos (\alpha-\beta)=\cos \alpha\cdot\cos \beta+\sin \alpha\cdot\sin \beta

Applying this formula to angles θ and φ, and substituting with the values found, we get:


\begin{gathered} \cos (\theta-\varphi)=\cos \theta\cdot\cos \varphi+\sin \theta\cdot\sin \varphi \\ \cos (\theta-\varphi)=\frac{2\sqrt[]{2}}{3}\cdot(8)/(17)+(1)/(3)\cdot(15)/(17) \\ \cos (\theta-\varphi)=\frac{16\sqrt[]{2}}{51}+(15)/(51) \\ \cos (\theta-\varphi)=\frac{16\sqrt[]{2}+15}{51} \end{gathered}

User Timothy Clotworthy
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