If - lies in the first quadrant and cos-817- then the value of cos30o-cos 45o-cos 120o- is

The correct option is A (√(3) 12+1√(2) )2317 cosθ = #10;8 17 ⇒sinθ = 15 17 #10;Now, #10; cos( 30+θ #160;) +cos( #10;45 θ #160;) +c ...

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Byju's Answer

Standard XII

Mathematics

Sin(A+B)Sin(A-B)

If θ lies i...

Question

If $θ$ lies in the first quadrant and $cos θ = \frac{8}{17}$ , then the value of $cos (30^{o} + θ) + cos (45^{o} - θ) + cos (120^{o} - θ)$ is

(\frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}}) \frac{23}{17}

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(\frac{\sqrt{3} + 1}{2} + \frac{1}{\sqrt{2}}) \frac{23}{17}

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(\frac{\sqrt{3} - 1}{2} - \frac{1}{\sqrt{2}}) \frac{23}{17}

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(\frac{\sqrt{3} + 1}{2} - \frac{1}{\sqrt{2}}) \frac{23}{17}

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Solution

The correct option is A $(\frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}}) \frac{23}{17}$
$cos θ = \frac{8}{17} \Rightarrow sin θ = \frac{15}{17}$
Now,
$cos (30 + θ) + cos (45 - θ) + cos (120 - θ) = cos (30 + θ) + cos (45 - θ) + cos (60 + θ) = cos 30 cos θ - sin 30 sin θ + cos 45 cos θ + sin 45 sin θ + cos 60 cos θ - sin 60 sin θ = \frac{\sqrt{3}}{2} . \frac{8}{17} - \frac{1}{2} . \frac{15}{17} + \frac{1}{\sqrt{2}} . \frac{8}{17} + \frac{1}{\sqrt{2}} . \frac{15}{17} + \frac{1}{2} . \frac{8}{17} - \frac{\sqrt{3}}{2} \frac{15}{17} = \frac{23}{17} (\frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}})$

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If θ lies in the first quadrant and cosθ=817, then the value of cos(30o+θ)+cos(45o−θ)+cos(120o−θ) is

The correct option is A (√3−12+1√2)2317cosθ=817⇒sinθ=1517 Now, cos(30+θ)+cos(45−θ)+cos(120−θ)=cos(30+θ)+cos(45−θ)+cos(60+θ)=cos30cosθ−sin30sinθ+cos45cosθ+sin45sinθ+cos60cosθ−sin60sinθ=√32.817−12.1517+1√2.817+1√2.1517+12.817−√321517=2317(√3−12+1√2)

If $θ$ lies in the first quadrant and $cos θ = \frac{8}{17}$ , then the value of $cos (30^{o} + θ) + cos (45^{o} - θ) + cos (120^{o} - θ)$ is