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यदि `sin(pi cos theta)=cos(pi sin theta)` तो सिद्ध करें की
(i) `cos (thetapm(pi)/(4))=(1)/(2sqrt(2)) " " (ii) sin 2 theta =pm(3)/(4)`

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प्रश्न से `sin (pi cos theta)= cos (pi sin theta)`
`:. sin (pi cos theta)= sin ((pi)/(2)pm sin theta)`
`:. pi cos theta=npi+(-1)^(n){(pi)/(2)pmsin theta}`
या `pi cos=n(-1)^(n)((1)/(2)pmsin theta)`
या `cos theta pm sin theta=n+(-1)^(n)(1)/(2)`
यह समीकरण `a cos theta+ b sin theta=c` के रूप में है इसलिए वास्तविक हल के लिए`| n +(-1)^(n)(1)/(2)| le sqrt(2) rArr n=0,1`
दोनों स्थितियों में
`cos theta bar+ sin theta =(1)/(2)" "....(1)`
(i)अब (ii) के दोनों पक्षों को `sqrt(2)` से भाग देने पर,
`(1)/(sqrt(2))cos theta bar+(1)/(sqrt(2))=(1)/(2sqrt(2))`
या `"cos"(pi)/(4) cos theta bar+ "sin"(pi)/(4) sin theta =(1)/(2sqrt(2))`
या `cos theta. "cos" (pi)/(4)bar+sin theta. "sin"(pi)/(2)=(1)/(2sqrt(2))`
या `cos (theta pm(pi)/(4))=(1)/(2sqrt(2))" "....(2)`
(ii) पुनः `(cos theta bar+ sin theta)^(2)=((1)/(2))^(2)` या `cos^(2) theta+sin^(2)bar+ 2 cos theta. Sin theta=(1)/(4)`
या `1bar+sin 2 theta=(1)/(4)` या `pm2 theta=1-(1)/(4)=(3)/(4) :. Sin 2 theta=pm(3)/(4)`

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