Math, asked by sharanya37, 10 months ago

If x= sin 130° + cos130° then what is the value of x ?​

Answers

Answered by Yashbhoir25
14

Answer:

Step-by-step explanation:

X= Sin130 + Cos130 =Sin(180-50) + Cos(180-50)

=Sin50 - Cos50

=Sin50 - Cos(90-40)

=Sin50 - Sin40

=Sin(45+5) - Sin(45-5)

=2Cos45.Sin5

Since, the value of 2cos45.sin5 is positive

the value of X is also positive.

Answered by pulakmath007
2

If x = sin 130° + cos 130° then value x > 0

Given :

x = sin 130° + cos 130°

To find :

The value of x

Solution :

Step 1 of 2 :

Write down the given expression

Here the given expression is

x = sin 130° + cos 130°

Step 2 of 2 :

Find the value of the expression

\displaystyle \sf{  x = sin  \: {130}^{ \circ} +  cos  \: {130}^{ \circ}}

\displaystyle \sf \implies x = sin  \:( {180}^{ \circ}  - {50}^{ \circ} )+  cos  \: ({90}^{ \circ} + {40}^{ \circ})

\displaystyle \sf{ \implies x = sin  \: {50}^{ \circ}  - sin  \:  {40}^{ \circ}}

\displaystyle \sf{ \implies x = 2cos  \:  \bigg( \frac{ {50}^{ \circ} +  {40}^{ \circ}}{2} \bigg) sin  \: \bigg( \frac{ {50}^{ \circ} +  {40}^{ \circ}}{2} \bigg)}

\displaystyle \sf{ \implies x = 2cos  \:  \bigg( \frac{ {90}^{ \circ}}{2} \bigg) sin   \bigg( \frac{ {10}^{ \circ}}{2} \bigg)}

\displaystyle \sf{ \implies x = 2cos  \: {45}^{ \circ}   sin  \:  {5}^{ \circ}}

\displaystyle \sf{ \implies x = 2   \times  \frac{1}{ \sqrt{2} } \times    sin  \:  {5}^{ \circ}}

\displaystyle \sf{ \implies x =  \sqrt{2}    \:  \:   sin  \:  {5}^{ \circ}}

Now 5° lies in first quadrant

∴ sin 5° is positive

∴ x > 0

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