Answer to Question #110558 in Trigonometry for atta

Question

1)prove the identity (sin theta +cos theta)(1 - sin theta cos theta)= sin cube theta + cos
cube theta
2)hence solve the equation (sin theta + cos theta)(1- sin theta cos theta)=3 cos cube theta for 0 degree <=theta<=360 degree

Answer 1

Given : $(\sin\theta+\cos\theta)(1-\sin\theta\cos\theta)=\sin^3\theta+\cos^3\theta$

Important:

• Since this is a very basic identity starting on either side is not a big deal but try to identify the easiest side to start.
• Working out L.H.S to R.H.S needs the application of the identity $\small\cos^2\theta+\sin^2\theta =\small 1$ & some simplifications whereas R.H.S to L.H.S needs $a^3+b^3=(a+b)(a^2-ab+b^2)$ which gives the answer in seconds.
• Division by $\cos^3\theta$ is possible because the value for theta when $\cos\theta=0$ is not a solution for the equation in the 2nd part.

Calculations

1).$\qquad \begin{aligned} \small\sin^3\theta+\cos^3\theta&=\small(\sin\theta+\cos\theta)(\sin^2\theta-\sin\theta\cos\theta+\cos^2\theta)\\ &=\small(\sin\theta+\cos\theta)(1-\sin\theta\cos\theta)\cdots\cdots\cdots(\because\sin^2\theta+\cos^2\theta=1)\\ \end{aligned}$

Therefore, the above identity is proved.

2). Now it can be written that, $\small\sin^3\theta+\cos^3\theta = \small 3\cos^3\theta$

$\qquad \begin{aligned} \therefore\,\small \sin^3\theta&= \small 2\cos^3\theta\\ \small \tan^3\theta& = \small 2 \cdots\cdots (division \,by\, \cos^3\theta) \\ \small \tan\theta&= \small 2^{\frac{1}{3}}\\ \small \text{but,} \, \tan\theta&= \small \tan(180+\theta)=\small 2^{\frac{1}{3}}\cdots\cdots(\because\, 0\degree<=\theta<=360\degree)\\ \small \therefore \theta&= \small \tan^{-1}(2^\frac{1}{3})\\ &= \small \bold{51.56\degree}\\ \small 180\degree+\theta &= \small \bold{231.56\degree}\\ \end{aligned}$