NIMCET 2016 Question Paper with solutions

Question : 98 of 120

Marks: +1, -0

The value of cos20° + cos 100° + cos 140° is;

Solution:

\cos20^{∘}+\cos100^{∘}+\cos140^{∘}

⇒(\cos20^{∘}+\cos100^{∘})+\cos140^{∘}

=2\cos(\;\;{20^{∘}+100^{∘}}/{2})\cos(\;\;{20^{∘}-100^{∘}}/{2})+\cos140^{∘}

[∵\cosC+\cosD=2\cos\;\;{C+D}/{2}\cos\;\;{C-D}/{2}]

=2\cos60^{∘}\cos(-40^{∘})+\cos140^{∘}

=2\cos60^{∘}\cos40^{∘}+\cos140^{∘}

[∵\cos(-θ)=\cosθ]

=2×\;\;{1}/{2}\cos40^{∘}+\cos140^{∘}

=2\cos(\;\;{140^{∘}+40^{∘}}/{2})\cos(\;\;{140^{∘}-40^{∘}}/{2})

=2\cos90^{∘}⋅\cos50^{∘}

[∵\cos90°=0]

⇒0

Go to Question:

Prev Question

Next Question