Given $A+B=135^\circ$ ,i have to find range of values of $\sin^2A+\sin^2B$

Question

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 $0 <A< 135$\

$0 <2A< 270$

$45 <2A+45< 315$

$\cos45 <cos(2A+45)< \cos315$

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

I am stuck here as to where i have gone wrong

Thanks

Okay mention that you are looking for inequality in the question. — N.S.JOHN, Feb 07 '17 at 08:22
@KanwaljitSingh, See http://math.stackexchange.com/questions/345703/prove-that-cos-a-b-cos-a-b-cos-2a-sin-2b — lab bhattacharjee, Feb 07 '17 at 13:30

Christian Blatter · Accepted Answer · 2017-02-08T07:41:35.610

2

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 $0<A<{3\pi\over4}$, and $\phi:=2A+{\pi\over4}$ satisfies ${\pi\over4}<\phi<{7\pi\over4}$, hence $-1\leq\cos\phi<{1\over\sqrt{2}}$. (Look at the graph of $\cos$ in the interval ${\pi\over4}<\phi<{7\pi\over4}$.) It follows that $Q=1-{1\over\sqrt{2}}\cos\phi$ satisfies $${1\over2}<Q\leq1+{1\over\sqrt{2}}\ .$$ Your line $\cos45^\circ <\cos(2A+45^\circ)< \cos315^\circ$ is wrong.

edited Feb 08 '17 at 07:41

answered Feb 07 '17 at 11:09

Christian Blatter

226,825

Sir, in the question it is given that A,B are angles of a triangle ABC – Taylor Ted Feb 07 '17 at 11:11
@TaylorTed, There is no information in your question that $A,B$ are angles of a triangle. – Alex Silva Feb 07 '17 at 11:28
Yes sorry my bad, i will edit it – Taylor Ted Feb 07 '17 at 11:34
@Christian Blatter Why i cannot take cosine in inequality $pi/4<phi<7pi/4$ – Taylor Ted Feb 08 '17 at 04:43
i mean after finding phi , how are you finding cos phi – Taylor Ted Feb 08 '17 at 04:44
@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 $$\cos(\frac\pi4)>\cos(\frac\pi2)$$ – woogie Feb 08 '17 at 08:19

score 1 · Answer 2 · answered Feb 07 '17 at 08:56

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

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

answered Feb 07 '17 at 08:56

meta_warrior

3,288
18
34

I don't know about differentiation. btw what is wrong with my solution – Taylor Ted Feb 07 '17 at 09:24
You are using the inequality of domain to indicate directly the inequality of the image – Zelos Malum Feb 07 '17 at 11:13
@ZelosMalum ok thanks i see it – Taylor Ted Feb 08 '17 at 08:20

Given $A+B=135^\circ$ ,i have to find range of values of $\sin^2A+\sin^2B$

2 Answers2