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Prove that(i) sin80° cos20° - cos80° sin20° = √3/2(ii) cos45° cos15° - sin45° sin15° = 1/2(iii) cos75° cos15° + sin75° sin15° = 1/2(iv) sin40° cos20° + cos40° sin20° = √3/2(v) cos130° cos40° + sin130° sin40° = 0 |
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Answer» (i) sin80°cos20° - cos80°sin20° = sin(80° - 20°) (using sin(A - B) = sinAcosB - cosAsinB) = sin60° = \(\frac{\sqrt{3}}2{}\) (ii) cos45°cos15° - sin45°sin15° = cos(45° + 15°) (Using cos(A + B) = cosAcosB - sinAsinB) = cos60° = \(\frac{1}{2}\) (iii) cos75°cos15° + sin75°sin15° = cos(75° - 15°) (using cos(A - B) = cosAcosB + sinAsinB) = cos60° = \(\frac{1}{2}\) (iv) sin40°cos20° + cos40°sin20° = sin(40° + 20°) (using sin(A + B) = sinAcosB + cosAsinB) = sin60° (v) cos130°cos40° + sin130°sin40° = cos(130° - 40°) (using cos(A - B) = cosAcosB + sinAsinB) = cos90° = 0 |
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