Maths Questions

Solution : We have, L.H.S = cos A cos (60 – A) cos (60 + A) $\implies$ L.H.S = cos A ($cos^2 60 – sin^2 A$) [ By using this formula, cos (A + B) cos (A – B) = $cos^2 A$ – $sin^2 B$  above ] $\implies$ L.H.S = cos A ($1\over 4$ – […]

Solution : We have, L.H.S = sin A sin (60 – A) sin (60 + A) $\implies$ L.H.S = sin A ($sin^2 60 – sin^2 A$) [ By using this formula, sin (A + B) sin (A – B) = $sin^2 A$ – $sin^2 B$  above ] $\implies$ L.H.S = sin A ($3\over 4$ –

Solution : The value of tan 120 degrees is $-\sqrt{3}$. Proof : We Know that tan 60 = $\sqrt{3}$. By using formula of tan 2A, tan 120 = $2 tan 60\over 1 – tan^2 60$ = $2 \times \sqrt{3}\over 1 – 3$ $\implies$  tan 120 = $-\sqrt{3}$

Solution : The value of cos 120 degrees is $-1\over 2$. Proof : We Know that sin 60 = $\sqrt{3}\over 2$ and cos 60 = $1\over 2$ By using cos 2A formula, cos 120 = $cos^2 60$ – $sin^2 60$ = $1\over 4$ – $3\over 4$ $\implies$  cos 120 = $-1\over 2$

Solution : The value of sin 120 degrees is $\sqrt{3}\over 2$. Proof : We Know that sin 60 = $\sqrt{3}\over 2$ and cos 60 = $1\over 2$ By using the formula of sin 2A, sin 120 = 2 sin 60 cos 60 = 2 $\times$ $\sqrt{3}\over 2$ $\times$ $1\over 2$ $\implies$  sin 120 = \(\sqrt{3}\over

Solution : The formula of cos C – cos D is 2 sin($C + D\over 2$) sin($D – C\over 2$). Example : Express cos $6\theta$ – cos $8\theta$ as a product. Solution : We have, cos $6\theta$ – cos $8\theta$ = 2 sin($6\theta + 8\theta\over 2$) sin($8\theta – 6 \theta\over 2$) = 2 sin $7\theta$ sin

Solution : The formula of cos C + cos D is 2 cos($C + D\over 2$) cos($C – D\over 2$). Example : Express cos $4\theta$ + cos $8\theta$ as a product. Solution : We have, cos $4\theta$ + cos $8\theta$ = 2 cos($8\theta + 4\theta\over 2$) cos($8\theta – 4\theta\over 2$) = 2 cos $6\theta$ cos

Solution : The formula of sin C – sin D is 2 sin($C – D\over 2$) cos($C + D\over 2$). Example : Express sin $6\theta$ – sin $2\theta$ as a product. Solution : We have, sin $6\theta$ – sin $2\theta$ = 2 sin($6\theta – 2\theta\over 2$) cos($6\theta + 2\theta\over 2$) = 2 sin $2\theta$ cos

Solution : The formula of sin C + sin D is 2 sin($C + D\over 2$) cos($C – D\over 2$). Example : Express sin $4\theta$ + sin $2\theta$ as a product. Solution : We have, sin $4\theta$ + sin $2\theta$ = 2 sin($4\theta + 2\theta\over 2$) cos($4\theta – 2\theta\over 2$) = 2 sin $3\theta$ cos

Solution : The formula of 2 sin A sin B is cos (A – B) – cos (A + B). Example : Solve : 2 sin 75 sin 15. Solution : We have, 2 sin 75 sin 15 = cos (75 – 15) – cos (75 + 15) = cos 60 – cos 90 =