Question

Prove :cos40 - sin 40=root 2 cos 5

Answer 1

Answer:

write sin 40⁰ as cos 50⁰ and apply cosC -cosD=[2sin{(C+D)/2}sin{D-C)/2}].Which will give u %√3sin10⁰ and sin10⁰=cos80⁰. hence proved

Answer 2

cos 40° - sin 40° = √2 cos 5° is proved

Given :

cos 40° - sin 40° = √2 cos 5°

To find :

To prove the expression

Formula :

We are aware of the Trigonometric identity that

$\displaystyle \sf cos \: C + cos \: D =2 \: cos \frac{C + D}{2} \: cos \frac{C - D}{2}$

Solution :

Step 1 of 2 :

Write down the given expression

The given expression is

cos 40° - sin 40° = √2 cos 5°

Step 2 of 2 :

Prove the expression

LHS

$= \sf cos {40}^{ \circ} + sin {40}^{ \circ}$

$= \sf cos {40}^{ \circ} + sin({90}^{ \circ} - {50}^{ \circ} )$

$= \sf cos {40}^{ \circ} + cos {50}^{ \circ}$

$\displaystyle \sf = 2 \: cos \frac{ {40}^{ \circ} + {50}^{ \circ} }{2} \: \: cos \frac{ {40}^{ \circ} - {50}^{ \circ} }{2}$

$\sf = 2 \: cos {45}^{ \circ} cos ( - {5}^{ \circ} \: )$

$\sf =2 \: cos {45}^{ \circ} cos {5}^{ \circ} \:$

$\displaystyle \sf =2 \: \times \frac{1}{ \sqrt{2} } \times cos {5}^{ \circ} \:$

$\displaystyle \sf = \sqrt{2} cos {5}^{ \circ} \:$

Hence proved

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