1.

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

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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