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Question:

$$\frac{2-3\sin\theta+\sin^3\theta}{\sin\theta+2}=2\sin\theta (\sin\theta-1)+\cos^2\theta$$


I don't know how to start with these problem. Normally these type of proof confuse me. In my book there are about 25 proof in which we have to prove L.H.S.=R.H.S out of there are 6 proof which i can't solve.

So was thinking about other way, and i remember that there are three methods to solve the proof (direct method, method of contrapositive and our normal method).

Can i use direct method to solve trigonometric proof(basically proof in which i have problem), means by assuming $\theta= \frac{\pi}{3}$ That would make stuff easy. But haven't seen any books using direct method to solve the trigonometric proof.

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  • $\begingroup$ Basically to prove that identity you have assume the LHS and use that to make look like the RHS through some algebraic manipulation. $\endgroup$
    – user60887
    Commented Aug 4, 2014 at 16:44
  • $\begingroup$ For theta = pi/3 is just veridication not proof $\endgroup$
    – DSinghvi
    Commented Aug 4, 2014 at 16:44
  • $\begingroup$ @DSinghvi So i can't use direct method to solve trigonometry proof! $\endgroup$
    – Freddy
    Commented Aug 4, 2014 at 17:00

3 Answers 3

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In this case the polynomial $x^3-3x+2$ has $x+2$ as a factor, so just divide and the rest should be easy.

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  • $\begingroup$ ok i got it, i didn't thought about polynomial. LOL $\endgroup$
    – Freddy
    Commented Aug 4, 2014 at 16:56
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Let $x=\sin\theta$, then: $$L.H.S.=\frac{2-3x+x^3}{x+2}=\frac{(x+2)(x-1)^2}{x+2}=(x-1)^2$$ $$R.H.S.=2x(x-1)+1-x^2=2x^2-2x+1-x^2=x^2-2x+1=(x-1)^2$$ So, $L.H.S.=R.H.S.=(\sin\theta-1)^2$

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Just see if you have done some algebra in the LHS you have factor of sinQ+2 of expression in numerator . and you almost get your RHS . Now just place Sin^2Q =1- Cos ^2Q

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