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Question Number 126575 by amns last updated on 22/Dec/20

$$\mathit{S}\mathit{o}\mathit{l}\mathit{v}\mathit{e}:sin\theta +cos\theta =1,\mathit{w}\mathit{h}\mathit{e}\mathit{n}0°⩽\theta ⩽90°$$

Commented by amns last updated on 22/Dec/20

$$Ineedahelptosolveit,\mathit{H}\mathit{e}\mathit{l}\mathit{p}\mathit{m}\mathit{e}.$$

Commented by benjo_mathlover last updated on 22/Dec/20

$$wrong.$$

Commented by amns last updated on 22/Dec/20

$$no...,how?$$

Commented by amns last updated on 22/Dec/20

$$itcanbe0°,30°,45°,60°,90°aswell$$

Commented by benjo_mathlover last updated on 22/Dec/20

$$youput\theta ={30}^{°}$$$$\sin \theta +\cos \theta =\frac{1}{2}+\frac{1}{2}\sqrt{3}\ne 1$$$$but\sin {}^2\theta +\cos {}^2\theta =1$$

Answered by MJS_new last updated on 22/Dec/20

$$s+\sqrt{1-s^2}=1$$$$\sqrt{1-s^2}=1-s$$$$1-s^2=1-2s+s^2$$$$2s^2-2s=0$$$$s(1-s)=0$$$$\sin \theta =0\vee \sin \theta =1\Rightarrow \theta =0°\vee \theta =90°$$

Answered by amns last updated on 22/Dec/20

$$benjo\mathrm{\_}mathlover,theans.is\theta =0°,90°$$$$hahaha...$$

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