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Given in a triangle $ABC$, $A+B=135^\circ$ ,i have to find value of $\sin^2A+\sin^2B$

ATTEMPT

i write as

$\sin^2A+\sin^2(135^\circ-A)$

$\sin^2A+\sin^2(45+A)$

$1-cos^2(A)+\sin^2(45+A)$

$1 - (cos(45+2A)(cos45))$

$1- \frac{1}{\sqrt2}cos(45+2A)$

Now $0

$0 <2A< 270$

$45 <2A+45< 315$

$\cos45

$\frac{1}{\sqrt2}

I am stuck here as to where i have gone wrong

Thanks

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    Why do you think there is going to be only one value? – 2017-02-07
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    Okay mention that you are looking for inequality in the question. – 2017-02-07
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    What you are doing in fourth step? – 2017-02-07
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    @KanwaljitSingh using $cos^2a-sin^2b=cos(a+b)cos(a-b)$ – 2017-02-07
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    @KanwaljitSingh, See http://math.stackexchange.com/questions/345703/prove-that-cos-a-b-cos-a-b-cos-2a-sin-2b – 2017-02-07

2 Answers 2

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Your $$Q=1-{1\over\sqrt{2}}\cos\left(2A+{\pi\over4}\right)$$ is correct. But in the formulation of the problem there were no further restrictions on $A$ and $B$. If $A$ and $B$ are supposed to be angles of a triangle then $0

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    Sir, in the question it is given that A,B are angles of a triangle ABC – 2017-02-07
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    @TaylorTed, There is no information in your question that $A,B$ are angles of a triangle. – 2017-02-07
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    Yes sorry my bad, i will edit it – 2017-02-07
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    @Christian Blatter Why i cannot take cosine in inequality $pi/4 – 2017-02-08
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    i mean after finding phi , how are you finding cos phi – 2017-02-08
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    @TaylorTed You can't take cosine of the inequality because cosine isn't an increasing function. Sometimes it goes up, sometimes it goes down. Example where it fails: $$\frac\pi4 < \frac\pi2$$ but $$\cos(\frac\pi4)>\cos(\frac\pi2)$$ – 2017-02-08
1

So you have $\sin^2A+\sin^2(135^\circ-A)=\sin^2A+\sin^2(45+A)$ At maximum and minimum values of this expression, we have

$$\frac{\text{d}}{\text{d}A}(\sin^2A+\sin^2(45^{\circ}+A))=0$$ $$2\sin(A)\cos(A)+2\sin(A+45^{\circ})\cos(A+45^{\circ})=0$$ $$\sin(2A)+\sin(2A+90^{\circ})=0$$ $$\sin(2A)+\cos(2A)=0$$ $$\sqrt{2}\sin(2A + 45^{\circ}) =0$$ ...

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    I don't know about differentiation. btw what is wrong with my solution – 2017-02-07
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    You are using the inequality of domain to indicate directly the inequality of the image – 2017-02-07
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    @ZelosMalum ok thanks i see it – 2017-02-08