Question

The value of cos² 45° - sin² 15° is​

Answer 1

Answer:

2 + √3/4

Step-by-step explanation:

To Find :-

Value of  cos² 45° - sin² 15°

Solution :-

cos² 45° - sin² 15°

= (cos45°)² - (sin15°)²

[∴ cos²A = (cosA)²]

We know that :-

• cos45° = 1/√2
• sin15° = √3 - 1/2√2

$\implies \left(\dfrac{1}{\sqrt{2}}\right)^2-\left(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)^2$

$=\dfrac{1}{2}-\dfrac{(\sqrt{3}-1)^2}{4(2)}$

$=\dfrac{1}{2}-\dfrac{3-2\sqrt{3}+1}{8}$

[∴ (a-b)² = a² - 2ab + b²]

$=\dfrac{1}{2}-\dfrac{4-2\sqrt{3}}{8}$

$=\dfrac{2(4)-(4-2\sqrt{3})}{8}$

$=\dfrac{8-4+2\sqrt{3}}{8}$

$=\dfrac{4+2\sqrt{3}}{8}$

$=\dfrac{2(2+\sqrt{3})}{8}$

$=\dfrac{2+\sqrt{3}}{4}$

cos² 45° - sin² 15° = 2 + √3/4

Know More :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$