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Integration by Parts

cos2 42∘-sin2...

Question

${cos}^{2} 42^{\circ} - {sin}^{2} 48^{\circ}$

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Solution

${cos}^{2} 42^{\circ} - {sin}^{2} 48^{\circ}$
= ${cos}^{2} (90^{\circ} - 48^{\circ}) - {sin}^{2} 48^{\circ}$
= ${sin}^{2} 48^{\circ} - {sin}^{2} 48^{\circ}$
= $0$

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Similar questions

3 - {sin}^{2} 48^{\circ} + {cos}^{2} 42^{\circ}

Q. Find the value of

{sin}^{2} 48^{\circ} - {cos}^{2} 12^{\circ}

{sin}^{2} 12^{\circ} + {sin}^{2} 21^{\circ} + {sin}^{2} 39^{\circ} + {sin}^{2} 48^{\circ} - {sin}^{2} 9^{\circ} - {sin}^{2} 18^{\circ} = 1

Q. Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 60° − sec 30° = 0
(iv) cot 34° − tan56° = 0
(v) sin²48° + sin²42° = 1
(vi) cos²72° + cos²18° = 1

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