Question

If the value of $\sin \alpha =\dfrac{3}{5}$ then find the value of

& \left( i \right)\sin 3\alpha \\\ & \left( ii \right)\cos 3\alpha \\\ & \left( iii \right)\tan 3\alpha \\\ \end{aligned}$$

Answer 1

We will solve $\left( i \right)\sin 3\alpha$ by using the formula $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta$ and putting $\theta =3\alpha$ and using $\sin \alpha =\dfrac{3}{5}\text{ and }{{\sin }^{3}}\alpha =\dfrac{27}{125}$, to compute $\cos 3\alpha =\cos \theta$ we will use $\cos \theta =\dfrac{\text{Base}}{\text{Hypotenuse}}$ and finally to compute ... to get result.

Complete step-by-step answer:
Let us solve this part (i),
We are given value of $\sin \alpha =\dfrac{3}{5}$
We have a trigonometric formula as $\sin 3\alpha =3\sin \alpha -4{{\sin }^{3}}\alpha$ as $\sin \alpha =\dfrac{3}{5}$ then cubing both sides we have:
$\Rightarrow \sin 3\alpha ={{\left( \dfrac{3}{5} \right)}^{3}}=\dfrac{27}{125}$
So, by applying above stated formula and substituting values, we get:

& \Rightarrow \sin 3\alpha =3\sin \alpha -4{{\sin }^{3}}\alpha \\\ & \Rightarrow \sin 3\alpha =3\left( \dfrac{3}{5} \right)-4\left( \dfrac{27}{125} \right) \\\ & \Rightarrow \sin 3\alpha =\dfrac{9}{5}-\dfrac{108}{125} \\\ & \Rightarrow \sin 3\alpha =\dfrac{25\times 9-108}{125} \\\ & \Rightarrow \sin 3\alpha =\dfrac{117}{125} \\\ \end{aligned}$$ Therefore, the value of $$\left( i \right)\sin 3\alpha =\dfrac{117}{125}$$ Consider part (ii) we have, Let $3\alpha =\theta $ then $\sin \theta =\dfrac{117}{125}$ and we have to compute $\cos \theta $ Consider triangle ABC as below with $\angle C\text{ as }\theta $  Then, as $$\sin \theta =\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\text{ and }\sin \theta =\dfrac{117}{125}$$ Perpendicular AB = 117 and Hypotenuse AC = 125. Now, we will compute side BC of triangle ABC by using Pythagoras theorem stated as below: In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of other two sides. In $\Delta ABC$ $$\begin{aligned} & {{\left( AC \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}} \\\ & {{\left( 125 \right)}^{2}}={{\left( 117 \right)}^{2}}+{{\left( BC \right)}^{2}} \\\ & {{\left( BC \right)}^{2}}={{\left( 125 \right)}^{2}}+{{\left( 117 \right)}^{2}} \\\ & B{{C}^{2}}=15625-13689 \\\ & B{{C}^{2}}=1936 \\\ & BC=44 \\\ \end{aligned}$$ So length of side BC = 44. Now, we know that, $$\cos \theta =\dfrac{\text{Base}}{\text{Hypotenuse}}$$ $$\begin{aligned} & \cos \theta =\dfrac{\text{Base}\to \text{BC}}{\text{Hypotenuse}\to \text{AC}} \\\ & \cos \theta =\dfrac{\text{44}}{\text{125}} \\\ \end{aligned}$$ The value of $$\left( ii \right)\cos 3\alpha =\dfrac{44}{125}$$ Finally we compute $\tan \theta $ by using formula $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ $$\text{as }\theta =3\alpha \Rightarrow \tan 3\alpha =\dfrac{\sin 3\alpha }{\cos 3\alpha }$$ Substituting the values of $$\sin 3\alpha \text{ as }\dfrac{117}{125}\text{ and }\cos 3\alpha \text{ as }\dfrac{44}{125}$$ $$\begin{aligned} & \tan 3\alpha =\dfrac{\dfrac{117}{125}}{\dfrac{44}{125}} \\\ & \tan 3\alpha =\dfrac{117}{44} \\\ \end{aligned}$$ The value of $$\left( iii \right)\tan 3\alpha =\dfrac{117}{44}$$ Hence, our answer is: $$\begin{aligned} & \left( i \right)\sin 3\alpha =\dfrac{117}{125} \\\ & \left( ii \right)\cos 3\alpha =\dfrac{44}{125} \\\ & \left( iii \right)\tan 3\alpha =\dfrac{117}{44} \\\ \end{aligned}$$ **Note:** Another way to calculate value of $\cos 3\alpha $ from $\sin 3\alpha $ is by using the trigonometric formula $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ $$\begin{aligned} & \sin 3\alpha =\dfrac{117}{125} \\\ & \Rightarrow \sqrt{1-{{\cos }^{2}}3\alpha }=\dfrac{117}{125} \\\ & \Rightarrow 1-{{\cos }^{2}}3\alpha ={{\left( \dfrac{117}{125} \right)}^{2}}=\dfrac{13689}{15625} \\\ & \Rightarrow {{\cos }^{2}}3\alpha =1-\dfrac{13689}{15625}=\dfrac{15625-13689}{15625} \\\ & \Rightarrow \cos 3\alpha =\dfrac{44}{125} \\\ \end{aligned}$$ Which is the correct value.