Trigonometry Questions

Solution : The value of cos 54 degrees is $\sqrt{10 – 2\sqrt{5}}\over 4$. Proof : We know that value of sin 36 degrees is $\sqrt{10 – 2\sqrt{5}}\over 4$. Since 54 degree is the complement of 36 degree. $\therefore$ cos 54 = sin(90 – 36) = sin 36 = $\sqrt{10 – 2\sqrt{5}}\over 4$ Hence, cos 54 […]

Solution : The value of sin 54 degrees is $\sqrt{5} + 1\over 4$. Proof : We know that value of cos 36 degrees is $\sqrt{5} + 1\over 4$. Since 54 degree is the complement of 36 degree. $\therefore$ sin 54 = sin(90 – 36) = cos 36 = $\sqrt{5} + 1\over 4$ Hence, sin 54

Solution : The value of sin 36 degrees is $\sqrt{10 – 2\sqrt{5}}\over 4$. Proof : We know that the value of cos 36 degrees is $\sqrt{5} + 1\over 4$. Putting $\theta$ = 36 in  $cos \theta$ = $\sqrt{1 – sin^2 \theta}$, we get cos 36 = $\sqrt{1 – sin^2 36}$ = \(\sqrt{1 – ({\sqrt{5} +

Solution : The value of cos 36 degrees is $\sqrt{5} + 1\over 4$. Proof : We know that the value sin 18 degrees is $\sqrt{5} – 1\over 4$. We have, $cos 2\theta$ = $1 – 2 sin^2 \theta$ $\therefore$   cos 36 = 1 – $2 sin^2 18$ $\implies$  cos 36 = 1 – 2\(({\sqrt{5} –

Solution : The value of cos 72 degrees is $\sqrt{5} – 1\over 4$. Proof : We know that value of sin 18 degrees is $\sqrt{5} – 1\over 4$. Since 72 degree is the complement of 18 degree. $\therefore$ cos 72 = cos(90 – 18) = sin 18 = $\sqrt{5} – 1\over 4$ Hence, cos 72

Solution : The value of sin 72 degrees is $\sqrt{10 + 2\sqrt{5}}\over 4$. Proof : We know that value of cos 18 degrees is $\sqrt{10 + 2\sqrt{5}}\over 4$. Since 72 degree is the complement of 18 degree. $\therefore$ sin 72 = sin(90 – 18) = cos 18 = $\sqrt{10 + 2\sqrt{5}}\over 4$ Hence, sin 72

Solution : The value of cos 18 degrees is $\sqrt{10 + 2\sqrt{5}}\over 4$. Proof : We know that the value of sin 18 degrees is $\sqrt{5} – 1\over 4$. Putting $\theta$ = 18 in  $cos \theta$ = $\sqrt{1 – sin^2 \theta}$, we get cos 18 = $\sqrt{1 – sin^2 18}$ = \(\sqrt{1 – ({\sqrt{5} –

Solution : The value of sin 18 degrees is $\sqrt{5} – 1\over 4$. Proof :  Let $\theta$ = 18 degrees. Then, $5\theta$ = 90 degrees $\implies$   $2\theta$ + $3\theta$ = 90 $\implies$   $2\theta$ = 90 – $3\theta$ $\implies$   $sin 2\theta$ = $sin (90 – 3\theta)$ $\implies$  $sin 2\theta$ = $cos 3\theta$ By using the formula

Solution : We have, L.H.S = cot A + cot (60 + A) –  cot (60 – A) $\implies$ L.H.S = $1\over tan A$ + $1\over tan (60 + A)$ – $1\over tan (60 – A)$ $\implies$ L.H.S = $1\over tan A$+ $1 – \sqrt{3} tan A\over \sqrt{3} + tan A$ – \(1 + \sqrt{3}

Solution : We have, L.H.S = tan A + tan (60 + A) –  tan (60 – A) $\implies$ L.H.S = tan A + $\sqrt{3} + tan A\over 1 – \sqrt{3} tan A$ – $\sqrt{3} – tan A\over 1 + \sqrt{3} tan A$ [ By using this formula, tan (A + B) = \(tan A