Question

If sin square theta + sin theta = 1, then show that cos square theta + cos raised to 4 theta = 1, I will everyone who answers a thank you and 5 stars please help

Answer 1

Given that,


sin²∅ + sin∅ = 1

To prove :- cos²∅ + cos⁴∅ = 1

Proof :-


Here we are given

sin²∅ + sin∅ = 1

But we know that sin²∅ + cos²∅ = 1

So we can equate them

=> sin²∅ + sin∅ = sin²∅ + cos²∅

=> sin∅ = sin²∅ + cos²∅ - sin²∅

=> sin∅ = cos²∅


Now we know that, cos²∅ = sin∅

squaring both sides,

(cos²∅)² = (sin∅)²

=> cos⁴∅ = sin²∅

We have to show that cos²∅ + cos⁴∅ = 1

We have derived that, cos²∅ = sin∅ and cos⁴∅ = sin²∅

So substituting the value,

cos²∅ + cos⁴∅ = 1

=> sin∅ + sin²∅ = 1

=> 1 = 1 (since it's given that sin∅ + sin²∅ = 1)

L.H.S = R.H.S

Hence Proved :)

Answer 2

Given,

$\sin^{2}\theta+\sin\theta=1\\\;\\\sin^{2}\theta+2.\sin\theta.\frac{1}{2}=1\\\;\\\sin^{2}\theta+2.\sin\theta.\frac{1}{2}+(\frac{1}{2})^2-\frac{1}{4}=1\\\;\\(\sin\theta+\frac{1}{2})^2-\frac{1}{4}=1\\\;\\(\sin\theta+\frac{1}{2})^2=1+\frac{1}{4}\\\;\\(\sin\theta+\frac{1}{2})^2=\frac{4+1}{4}\\\;\\(\sin\theta+\frac{1}{2})^2=\frac{5}{4}\\\;\\(\sin\theta+\frac{1}{2})=\frac{\sqrt{5}}{2}\\\;\\\sin\theta=\frac{\sqrt{5}}{2}-\frac{1}{2}\\\;\\\sin\theta=\frac{\sqrt{5}-1}{2}\\\;\\\text{Making square of both sides}$

$\sin^2\theta=(\frac{\sqrt{5}-1}{2})^2\\\;\\\sin^2\theta=\frac{5+1-2\sqrt{5}}{4}\\\;\\\sin^2\theta=\frac{6-2\sqrt{5}}{4}\\\;\\\sin^2\theta=\frac{3-\sqrt{5}}{2}$

We know that,

$\sin^2\theta=1-\cos^2\theta$

$sin^2\theta=\frac{3-\sqrt{5}}{2}\\\;\\1-\cos^2\theta=\frac{3-\sqrt{5}}{2}\\\;\\\cos^\theta=1-(\frac{3-\sqrt{5}}{2})\\\;\\\cos^2\theta=\frac{2-3+\sqrt{5}}{2}\\\;\\cos^2\theta=\frac{\sqrt{5}-1}{2}...............i)\\\;\\\text{Again,Making square of both sides}\\\;\\\cos^4\theta=(\frac{\sqrt{5}-1}{2})^2\\\;\\\cos^4\theta=\frac{5+1-2\sqrt{5}}{4}\\\;\\\cos^4\theta=\frac{6-2\sqrt{5}}{4}\\\;\\\cos^4\theta=\frac{3-\sqrt{5}}{2}..............ii)\\\;\\\text{Adding equ i) and ii)}$

$cos^2\theta+\cos^4\theta=\frac{\sqrt{5}-1}{2}+\frac{3-\sqrt{5}}{2}\\\;\\cos^2\theta+\cos^4\theta=\frac{\sqrt{5}-1+3-\sqrt{5}}{2}\\\;\\cos^2\theta+\cos^4\theta=\frac{2}{2}\\\;\\cos^2\theta+\cos^4\theta=1\\\;\\\textbf{Hence proved.}$