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Question 5, Exercise 10.1
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\left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan\alpha$ is... arrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$ As \begin{align... \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan\alpha$ is... arrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$ As \begin{align
Question 3, Exercise 10.1
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actly $\cos \left( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ $0\le u\le \dfrac{\pi }... actly $\tan \left( u-v \right)$ ====Solution==== Given \begin{align}\sin u&=\dfrac{3}{5},\,\,\,\, 0\l... actly $\sin \left( u-v \right)$ ====Solution==== Given \begin{align}\sin u&=\dfrac{3}{5},\,\,\,\, 0\l... actly $\cos \left( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ $0\le u\le \dfrac{\pi }
Question, Exercise 10.1
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\left( \alpha -\beta \right)$. ====Solution==== Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd ... \left( \alpha +\beta \right)$. ====Solution==== Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd ... \left( \alpha +\beta \right)$ . ====Solution==== Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd
Question 2, Exercise 10.2
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ant, then find $\sin 2\theta $. ====Solution==== Given: $\sin \theta =\dfrac{5}{13}$. Using the identit... ant, then find $\cos 2\theta $. ====Solution==== Given: $\sin \theta =\dfrac{5}{13}$. Using the identit... ant, then find $\tan 2\theta $. ====Solution==== Given: $\sin \theta =\dfrac{5}{13}$. Using the identit
Question 4 and 5, Exercise 10.2
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find $\sin \dfrac{\theta }{2}$. ====Solution==== Given: $\cos \theta =-\dfrac{3}{7}$ and terminal ray of... exactly $\sin \dfrac{2\pi }{3}$. ====Solution==== Given: $\sin \dfrac{2\pi }{3}$.\\ By using double angle... exactly $\cos \dfrac{2\pi }{3}$. ====Solution==== Given: $\cos \dfrac{2\pi }{3}$ \\ By using double angle
Question 3, Exercise 10.2
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adrant, then find $\sin2\theta$. ====Solution==== Given: $\sin \theta =\dfrac{4}{5}$ Terminal ray of $\t... find $\cos \dfrac{\theta }{2}$. ====Solution==== Given: $\sin \theta =\dfrac{4}{5}$ Terminal ray of $\t
Question 1, Exercise 10.2
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in 2\theta ,\,\,\cos 2\theta$ and $\tan 2\theta$, given $\tan \theta =-\dfrac{1}{5}$, $\theta$ in quadran