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Grade 11Trigonometry

Please answer the question if your answer is correct then I will approve thanks

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Profile image of Chandan Kumar
8 Years agoGrade 11
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2 Answers

Profile image of Sandeep Bhamoo
8 Years ago
Sin(©) + cos(©) =√2sin(45°+©) =√2sin(a) , 45°+©=2nπ±a, ©=2nπ-45°±a which is second option don`t forget to approve it
Profile image of Soumendu Majumdar
ApprovedApproved Tutor Answer8 Years ago
Dear Student,
sin\theta + cos\theta = \sqrt{2}cos\alpha \Rightarrow 1/\sqrt{2}sin\Theta + 1/\sqrt{2}cos\Theta =cos\alpha
\Rightarrow cos(\Theta -45^{\circ}) =cos\alpha  since cos A cos B + sin A sin B = cos (A-B)  andsin 45^{\circ}=cos45^{\circ}= 1/\sqrt{2}
so according to general solutions of cos A = cos B
\theta- 45^{\circ}=2n\Pi +- \alpha
\Rightarrow \theta=2n\Pi +\Pi /4 + \alpha    where n\epsilon Z
Hope it helps!
with regards,
Soumendu
 

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