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Prove that: i) sin40^(@)cos50^(@)+cos4...

Prove that:
i) `sin40^(@)cos50^(@)+cos40^(@)sin50^(@)=1`
ii) `cot(270^(@)-theta)cot(270^(@)+theta)(cot(540^(@)-theta)cot(540^(@)+theta)=1`
iii) `cospi/8+cos(3pi)/8+cos(5pi)/8+cos(7pi)/8=0`

Text Solution

Verified by Experts

LHS `=sin40^(@)cos50^(@)+cos40^(@)sin50^(@)`
`=sin40^(@)cos(90^(@)-40^(@))+cos40^(@)sin(90^(@)-40^(@))`
`=sin40^(@)+sin40^(@)+cos40^(@)cos40^(@)`
`=sin^(2)40^(@)+cos^(2)40^(@)=1=`RHS. Hence proved
ii) LHS `=cot(270^(@)-theta) cot(270^(@)+theta)cot(540^(@)-theta)`
`=tantheta(-tantheta).cot(360^(@)+(180^(@)-theta). cot(360^(@)+(180^(@)+theta))`
`-tan^(2)theta.cot(180^(@)-theta).cot(180^(@)+theta)`
`-tan^(2)theta(-cottheta)(cottheta)`
`-1/(cot^(2)theta)(-cot^(2)theta)`
=1=RHS Hence Proved.
iii) LHS`=(cospi)/8 + cos(3pi)/8+cos(5pi)/8+cos(7pi)/8`
`=cospi/8+cos(3pi)/8 + cos(pi-(3pi)/8)+cos(pi-pi/8)`
`=cospi/8+cos(3pi)/8-cos(3pi)/8-cospi/8`
=0 = RHS Hence proved.
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