Proving trigonometry identities

chwala
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Homework Statement
see attached
Relevant Equations
trigonometry concept
I was just looking at the problem below: there may be several ways to prove the identity:

question:
1626491496310.png


Mark scheme solution:

1626491538005.png


My take:
we may also use ##sin^{2}x+cos^{2}x≡(sin x+ cos x)(sin x-cosx)##...
we end up with(##\frac 2 {\sqrt{2}}##cos ∅)(##\frac 2 {\sqrt{2}}##sin ∅)=##2 sin ∅ cos ∅ ##= ##sin 2∅##
 
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The formula
cos(\alpha+\beta)=\cos\alpha cos\beta - \sin \alpha \sin \beta
is helpful.
 
anuttarasammyak said:
The formula
cos(\alpha+\beta)=\cos\alpha cos\beta - \sin \alpha \sin \beta
is helpful.
True, that's what I used...together with the one for sine...
 
chwala said:
My take:
we may also use ##sin^{2}x+cos^{2}x≡(sin x+ cos x)(sin x-cosx)##...
No, that's incorrect.
##\sin^2(x) + \cos^2(x) = 1##, for all x.
If you meant ##\sin^2(x) - \cos^2{x}##, that that does factor as you show above.
chwala said:
we end up with(##\frac 2 {\sqrt{2}}##cos ∅)(##\frac 2 {\sqrt{2}}##sin ∅)=##2 sin ∅ cos ∅ ##= ##sin 2∅##
 
Mark44 said:
No, that's incorrect.
##\sin^2(x) + \cos^2(x) = 1##, for all x.
If you meant ##\sin^2(x) - \cos^2{x}##, that that does factor as you show above.
yeah, true...i made a slight mistake there...it should be the difference of two squares, thanks mark...i.e
##[sin(∅+45)+cos(∅+45)]⋅[sin(∅+45)-cos(∅+45)]##
##[(sin ∅cos 45+cos ∅sin 45)+(cos ∅cos 45-sin ∅sin 45)]⋅
[(sin ∅cos 45+cos ∅sin 45)-(cos ∅cos 45-sin ∅sin 45)]##
thanks Mark cheers.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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