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sdas [7]
3 years ago
7

cos theta = 4/5, 0˚< theta < 90˚ use the information given to find sin2theta, cos2theta, and tan 2theta

Mathematics
2 answers:
NikAS [45]3 years ago
5 0
First calculate \sin\theta.

\sin^2\theta+\cos^2\theta=1\\\\\sin^2\theta=1-\cos^2\theta\\\\\sin^2\theta=1-\left(\dfrac{4}{5}\right)^2\\\\\\\sin^2\theta=1-\dfrac{16}{25}\\\\\\\sin^2\theta=\dfrac{9}{25}\quad|\sqrt{(\dots)}\\\\\\\sin\theta=\sqrt{\dfrac{9}{25}}\\\\\\\boxed{\sin\theta=\dfrac{3}{5}}

We take positive value of sinθ because 0° < θ < 90°
Now we could calculate sin2θ, cos2θ and tan2θ:

\sin2\theta=2\sin\theta\cos\theta=2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5}=\dfrac{2\cdot3\cdot4}{5\cdot5}=\dfrac{24}{25}\\\\\\\cos2\theta=\cos^2\theta-\sin^2\theta=\left(\dfrac{4}{5}\right)^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}\\\\\\&#10;\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=\dfrac{\frac{24}{25}}{\frac{7}{25}}=\dfrac{24\cdot25}{7\cdot25}=\dfrac{24}{7}=3\dfrac{3}{7}
yarga [219]3 years ago
5 0

Answer:

D edge

Step-by-step explanation:

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\bf \cfrac{1+cot^2(\theta )}{1+csc(\theta )}=\cfrac{1}{sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\implies \cfrac{1+\frac{cos^2(\theta )}{sin^2(\theta )}}{1+\frac{1}{sin(\theta )}}\implies \cfrac{~~\frac{sin^2(\theta )+cos^2(\theta )}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}\implies \cfrac{~~\frac{1}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}

\bf \cfrac{1}{\underset{sin(\theta )}{~~\begin{matrix} sin^2(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}\cdot \cfrac{~~\begin{matrix} sin(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{sin(\theta )+1}\implies \cfrac{1}{sin^2(\theta )+sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\ne \cfrac{1}{sin(\theta )}~\hfill

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