Math, asked by shivamkuma9589, 1 month ago

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

Answers

Answered by Ladylaurel
19

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
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