Prove that $\alpha + \beta=\frac {\pi}{2}$ [closed]

Question

It is given that-

(1) $0<\alpha,\beta<90$.

(2) $\sin^2\alpha+\sin^ 2\beta=\sin(\alpha+\beta).$

Prove that $\alpha + \beta=\frac {\pi}{2}$

Answer 1

NOTE: it's answer to initial question (without squares), downvoters.

It's very strightforward. $$ \sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\ \sin(\alpha + \beta) = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}\\ \sin\alpha + \sin\beta = \sin(\alpha + \beta)\Longrightarrow \sin\frac{\alpha+\beta}{2} = 0\text{ or }\cos\frac{\alpha-\beta}{2} = \cos\frac{\alpha+\beta}{2} $$ In the first case $$ \sin\frac{\alpha+\beta}{2} = 0\Longrightarrow \alpha + \beta = \pi n,0 < \alpha + \beta < \pi; $$ there ares no solutions. If $\alpha,\beta \color{red}\le \pi/2$, then $\alpha + \beta = \pi$.

In the second case, $$\cos\frac{\alpha-\beta}{2} = \cos\frac{\alpha+\beta}{2} \Longrightarrow 2\sin\frac\alpha2\sin\frac\beta2 = 0, $$ and $\alpha=2\pi k$ or $\beta=2\pi m$, $k,m\in\mathbb Z$.

Anyway, your statement is false.

Answer 2

We have -

$\sin^2\alpha+\sin^2\beta=\sin(\alpha+\beta)$

$\implies \sin\alpha\cos\beta+cos\alpha\sin\beta=\sin^2\alpha+\sin^2\beta$

$\implies\displaystyle\sin\alpha(\cos\beta-\sin\alpha)+\sin\beta(\cos\alpha-\sin\beta)=0$

$\implies2 \sin\alpha\cdot \sin\left({\frac{\beta+\frac \pi2-\alpha}{2}}\right)\cdot \sin\left({\frac {\frac{\pi}{2}-\alpha-\beta}{2}}\right)+2\sin\beta\cdot \sin\left({\frac{\alpha+\frac{\pi}{2}-\beta}{2}}\right)\cdot \sin\left({\frac{\frac{\pi}{2}-\beta-\alpha}{2}}\right)=0$

$\implies\sin\alpha\cdot \sin\left(\frac{\pi}{4}+\frac{\beta-\alpha}{2}\right)\cdot \sin\left(\frac{\pi}{4}-\frac{\alpha+\beta}{2}\right)+\sin\beta\cdot \sin\left(\frac{\pi}{4}+\frac{\alpha-\beta}{2}\right)\cdot \sin\left(\frac{\pi}{4}-\frac{\alpha+\beta}{2}\right)=0$

$\implies\sin\left(\frac{\pi}{4}-\frac{\beta+\alpha}{2}\right)=0$

$\implies\frac{\pi}{4}-\frac{\alpha+\beta}{2}=0$

$\implies \alpha+\beta=\frac{\pi}{2}$

Answer 3

Equation $(2)$ says that $$\sin\alpha(\sin\alpha-\cos\beta)=\sin\beta(\cos\alpha-\sin\beta)\ .$$ With $\beta:={\pi\over2}-\beta'$ we therefore have $$\sin\alpha(\sin\alpha-\sin\beta')=\sin\beta(\cos\alpha-\cos\beta')\ .$$ When $\alpha\ne\beta'$ the two sides of the last equation have different signs.

Answer 4

Its given (by you) that ${sin}^2\alpha+{sin}^2\beta={sin}(\alpha+\beta) ..........eqn (1)$

But ${sin}^2\beta=1-{cos}^2\beta$

Slotting the identity into $eqn(1)$ gives: $${sin}^2\alpha+(1-{cos}^2\beta)={sin}(\alpha+\beta) $$ $$1-({cos}^2\beta-{sin}^2\alpha)={sin}(\alpha+\beta)$$ $$1-{cos}(\alpha+\beta)={sin}(\alpha+\beta)$$ $${sin}(\alpha+\beta)+{cos}(\alpha+\beta)=1$$ Let $(\alpha+\beta)=A$ Therefore $sinA+cosA=1$

Solving this further by squaring both sides gives: $${sin}^2A+{cos}^2A+2sinAcosA=1$$ $$2sinAcosA=0$$ Therefore $sinA=0$ or $cos A=0$

This gives $A=0$ or $A=\frac{\pi}{2}$

Recall that $(\alpha+\beta)=A$

Therefore $\alpha+\beta=0$ or $\alpha+\beta=\frac{\pi}{2}$

Answer 5

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

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

$$\implies\sin(A+B)+\cos(A+B)\cos(A-B)=1$$

Using Weierstrass substitution for $\sin(A+B),\cos(A+B)$ and setting $\tan\dfrac{A+B}2=t,$

$$[1+\cos(A-B)]t^2-2t+1-\cos(A-B)=0$$

Using Weierstrass substitution for $\cos(A-B),$

$$t=1,\tan^2\dfrac{A-B}2$$

If $\tan\dfrac{A+B}2=1,\dfrac{A+B}2=n\pi+\dfrac\pi2\iff A+B=?$ where $n$

But I'm not sure about $\tan\dfrac{A+B}2=\tan^2\dfrac{A-B}2$