Question

2.2.Fnd
the value of (Sin45° + Cos45°).​

Answer 1

Appropriate Question :-

Find the value of :-

$\large{\sf{(sin \: {45}^{\circ} + cos \: {45}^{\circ})}}$

Answer :-

• (sin 45° + cos 45°) = √2

Step-by-step explanation:

To Find :-

• The value of $\sf{(sin \: {45}^{\circ} + cos \: {45}^{\circ})}$

★ Solution

$\sf{ \longrightarrow \: (sin \: {45}^{\circ} + cos \: {45}^{\circ})}$

By using the trigonometric ratio of

[sin 45° = 1/√2], We get,

$\sf{ \longrightarrow \: \bigg( \dfrac{1}{ \sqrt{2}} + cos \: {45}^{\circ} \bigg)}$

By using the trigonometric ratio of

[cos 45° = 1/√2], We get,

$\sf{ \longrightarrow \: \bigg( \dfrac{1}{ \sqrt{2}} + \dfrac{1}{ \sqrt{2}} \bigg)}$

By simplifying them,

$\sf{ \longrightarrow \: \dfrac{2}{ \sqrt{2}}}$

By rationalising the denominator,

multiplying the denominator and numerator with [√2], We get,

$\sf{ \longrightarrow \: \dfrac{2}{ \sqrt{2}} \times \dfrac{ \sqrt{2}}{ \sqrt{2}}}$

$\sf{ \longrightarrow \: \dfrac{2 \times \sqrt{2}}{ \sqrt{2} \times \sqrt{2}}}$

$\sf{ \longrightarrow \: \dfrac{2 \sqrt{2}}{2}}$

By cancelling [2] from both numerator and denominator, We get,

$\sf{ \longrightarrow \: \dfrac{ \not{2} \sqrt{2}}{ \not{2}}}$

$\bf{ \longrightarrow \: \sqrt{2}}$

Hence,

$\sf{ \bigstar \: \: \: \: \: (sin \: {45}^{\circ} + cos \: {45}^{\circ}) = \sqrt{2}}$

Things to remember :-

• sin 45° = 1/√2
• cos 45° = 1/√2