Given. cos(A-B)= cosAcosB +sinAsinB. value of cos15
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cos (45-30)=cos45.cos30+sin45.sin30
=root3/2.1/root2+1/root 2.1/2
=(root3+1)/2root2
=root3/2.1/root2+1/root 2.1/2
=(root3+1)/2root2
abhi178:
thank you very much
Answered by
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Let us take A=45° & B=30°
cos(A-B) = cosAcosB + sinAsinB
cos (45°-30°) = cos 45° cos30°+ sin45° sin30°
cos 15° = (1/√2)(√3/2) + (1/√2)(1/2)
cos 15° = √3/2√2 +1/2√2
cos 15° = (√3+1)/(2√2)
cos(A-B) = cosAcosB + sinAsinB
cos (45°-30°) = cos 45° cos30°+ sin45° sin30°
cos 15° = (1/√2)(√3/2) + (1/√2)(1/2)
cos 15° = √3/2√2 +1/2√2
cos 15° = (√3+1)/(2√2)
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