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Question: a. Express \(15\sin \theta - 8\cos \theta \) in the form \(R\sin (\theta - \alpha )\), where \[R > 0...

a. Express 15sin⁡θ−8cos⁡θ15\sin \theta - 8\cos \theta in the form Rsin⁡(θ−α)R\sin (\theta - \alpha ), where R>0R > 0 and 0∘<α<90∘.0^\circ < \alpha < 90^\circ .
Give the value of α\alpha correct to 22 decimal places.
b. Hence solve the equation 15sin⁡θ−8cos⁡θ=1015\sin \theta - 8\cos \theta = 10 for 0∘<θ<360∘.0^\circ < \theta < 360^\circ .

Explanation

Solution

Hint : o solve this question, we will start with expressing the given expression 15sin⁡θ−8cos⁡θ15\sin \theta - 8\cos \theta in the form Rsin⁡(θ−α)R\sin (\theta - \alpha ), now we will construct the right-angled triangle, and then with the help of that figure we will get the value of α\alpha . Now in the b part, to find the value of θ\theta , we will use the expression of a part, and on putting the value, we will get the required answer.

Complete step-by-step answer :
a. We have been given an expression 15sin⁡θ−8cos⁡θ15\sin \theta - 8\cos \theta we need to express it in the form Rsin⁡(θ−α)R\sin (\theta - \alpha ), where R>0 and 0∘<α<90∘.0^\circ < \alpha < 90^\circ . And then we need to give the value of α\alpha correct to 2 decimal places.
So, we are given the expression =15sin⁡θ−8cos⁡θ= 15\sin \theta - 8\cos \theta
On multiplying and dividing whole expression by 17, we get
=17(15sin⁡θ17−8cos⁡θ17)= 17(\dfrac{{15\sin \theta }}{{17}} - \dfrac{{8\cos \theta }}{{17}})
= 17\sin (\theta - \alpha )$$$ \ldots .eq.\left( 1 \right)$$ So, we get the expression15\sin \theta - 8\cos \theta intheformin the formR\sin (\theta - \alpha ).Now,wewilldrawaright−angledtriangle,togetthevalueof. Now, we will draw a right-angled triangle, to get the value of \alpha .![](https://www.vedantu.com/question−sets/7d92980d−ddbe−415d−b287−69e39d40817a6438343885150982417.png) ![](https://www.vedantu.com/question-sets/7d92980d-ddbe-415d-b287-69e39d40817a6438343885150982417.png) \cos \alpha = \dfrac{{15}}{{17}} \sin \alpha = \dfrac{8}{{17}} \tan \alpha = \dfrac{8}{{15}} \alpha = ta{n^{ - 1}}\dfrac{8}{{15}} \Rightarrow \alpha = 28.07Thus,thevalueof Thus, the value of\alpha $ correct to 2 decimal places is 28.07.

b. Now, we need to solve the equation 15sin⁡θ−8cos⁡θ=1015\sin \theta - 8\cos \theta = 10 for 0∘<θ<360∘.0^\circ < \theta < 360^\circ .
So, the given equation is 15sin⁡θ−8cos⁡θ=1015\sin \theta - 8\cos \theta = 10
From eq.(1),eq.\left( 1 \right), we get
17sin⁡(θ−α)=10 sin⁡(θ−α)=1017 \begin{gathered} 17\sin (\theta - \alpha ) = 10 \\\ \sin (\theta - \alpha ) = \dfrac{{10}}{{17}} \\\ \end{gathered}
θ=α+sin−11017\theta = \alpha + si{n^{ - 1}}\dfrac{{10}}{{17}}
Now, on putting the value of α\alpha from a part and the value of sin−11017,si{n^{ - 1}}\dfrac{{10}}{{17}}, we get
θ=28.07+36.03\theta = 28.07 + 36.03
⇒θ=64∘\Rightarrow \theta = 64^\circ, 0∘<θ<360∘.0^\circ < \theta < 360^\circ .
Thus, solution of 15sin⁡θ−8cos⁡θ=1015\sin \theta - 8\cos \theta = 10 for 0∘<θ<360∘,0^\circ < \theta < 360^\circ ,is 64∘.64^\circ .

Note : Students should note that, in the solution, we have put the value of tan−1815ta{n^{ - 1}}\dfrac{8}{{15}} and sin−11017,si{n^{ - 1}}\dfrac{{10}}{{17}}, to get these inverse values, we took the help of a calculator. Though there is a long method to solve inverses without calculators, that you will learn in higher classes.