Solveeit Logo
Login

Question

Question: Find the maximum and minimum values of the trigonometric expression \(5\cos \theta +3\sin \left( \df...

Find the maximum and minimum values of the trigonometric expression 5cos⁡θ+3sin⁡(π6−θ)+45\cos \theta +3\sin \left( \dfrac{\pi }{6}-\theta \right)+4

Explanation

Solution

Hint: Start by using the formula of sin(A-B). Then divide and multiply the expression given in the question by 7 and take sin⁡α=1314\sin \alpha =\dfrac{13}{14} . Also, use the fact that the range of the sine function is [-1,1] and is defined for all real numbers.

Complete step-by-step answer:
Now we will start with the simplification of the expression that is given in the question.
5cos⁡θ+3sin⁡(π6−θ)+45\cos \theta +3\sin \left( \dfrac{\pi }{6}-\theta \right)+4
Using the formula sin(A-B)=sinAcosB-cosAsinB.
5cos⁡θ+3sin⁡π6cos⁡θ−3cos⁡π6sin⁡θ+45\cos \theta +3\sin \dfrac{\pi }{6}\cos \theta -3\cos \dfrac{\pi }{6}\sin \theta +4
We know that sin⁡π6=12 and cosπ6=32\sin \dfrac{\pi }{6}=\dfrac{1}{2}\text{ and cos}\dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2} . So, our expression becomes:
5cos⁡θ+32cos⁡θ−332sin⁡θ+45\cos \theta +\dfrac{3}{2}\cos \theta -\dfrac{3\sqrt{3}}{2}\sin \theta +4
132cos⁡θ−332sin⁡θ+4\dfrac{13}{2}\cos \theta -\dfrac{3\sqrt{3}}{2}\sin \theta +4
Now we divide and multiply the expression by 7. On doing so, we get
14(1314cos⁡θ−3314sin⁡θ)+414\left( \dfrac{13}{14}\cos \theta -\dfrac{3\sqrt{3}}{14}\sin \theta \right)+4
We take 1314\dfrac{13}{14} to be equal to sinαsin\alpha , then we cos⁡α\cos \alpha can be calculated as:
sin⁡2α+cos⁡2α=1 ⇒cos⁡α=1−sin2α=1−169196=3314 \begin{aligned} & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1 \\\ & \Rightarrow \cos \alpha =\sqrt{1-si{{n}^{2}}\alpha }=\sqrt{1-\dfrac{169}{196}}=\dfrac{3\sqrt{3}}{14} \\\ \end{aligned}
Therefore, using the above assumption and result in the expression 14(1314cos⁡θ−3314sin⁡θ)+414\left( \dfrac{13}{14}\cos \theta -\dfrac{3\sqrt{3}}{14}\sin \theta \right)+4 , we get
14(1314cos⁡θ−3314sin⁡θ)+414\left( \dfrac{13}{14}\cos \theta -\dfrac{3\sqrt{3}}{14}\sin \theta \right)+4
=14(cosθsinα−sin⁡θcos⁡α)+4=14\left( cos\theta sin\alpha -\sin \theta \cos \alpha \right)+4
Now using the formula sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B , our expression becomes:
=14sin⁡(α−θ)+4=14\sin \left( \alpha -\theta \right)+4
Now we let α−θ=β\alpha -\theta =\beta . Therefore, we get our final expression to be 14sin⁡(α−θ)+414\sin \left( \alpha -\theta \right)+4 .
We know that the sine function can have a maximum value of 1 and minimum value of -1. Also, our final expression is maximum when sin⁡β\sin \beta is maximum and minimum when sin⁡β\sin \beta is minimum. The expression is minimum.
Therefore, the maximum and minimum value of the expression 5cos⁡θ+3sin⁡(π6−θ)+45\cos \theta +3\sin \left( \dfrac{\pi }{6}-\theta \right)+4 is 18 and -10, respectively.

Note: If you want, you can directly remember that the maximum and minimum value of the expression asin⁡θ−bcos⁡θa\sin \theta -b\cos \theta is a2+b2 and −a2+b2\sqrt{{{a}^{2}}+{{b}^{2}}}\text{ and }-\sqrt{{{a}^{2}}+{{b}^{2}}} , respectively. Also, for solving the above question, you can use the method of derivative, but that would be difficult to solve and would require a good hold on the concepts of inverse trigonometric functions.