Question

The value of sin 105° + cos 105° is :​

Answer 1

Given Question : -

• The value of sin 105° + cos 105° is ______

Answer

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$

$(1). \: \boxed{ \bf \: sin(90 \degree \:+ \: x) = cosx }$

$(2). \: \boxed{ \bf \: cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg) cos\bigg( \dfrac{x - y}{2} \bigg) }$

$\large\underline{\bold{Solution-}}$

Consider,

$\rm :\longmapsto\:sin105\degree \: + cos105\degree \:$

$\rm :\longmapsto\:sin(90\degree \: + \: 15\degree \:) + cos105\degree \:$

$\rm :\longmapsto\:cos15\degree \: + cos105\degree \:$

$\rm :\longmapsto\:cos105\degree \: + cos15\degree \:$

$\rm :\longmapsto\:2cos\bigg( \dfrac{105\degree \: + 15\degree \:}{2} \bigg) cos\bigg( \dfrac{105\degree \: - 15\degree \:}{2} \bigg)$

$\rm :\longmapsto\:2cos\bigg( \dfrac{120\degree \:}{2} \bigg) cos\bigg(\dfrac{90\degree \:}{2} \bigg)$

$\rm :\longmapsto\:2 \: cos60\degree \:sin45\degree \:$

$\rm :\longmapsto\:2 \: \times \dfrac{1}{2} \: \times \dfrac{1}{ \sqrt{2} }$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{2} }$

Additional Information :-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)

Answer 2

Answer:

Step-by-step explanation:

eya frnd...

Here is ur answer...

sin(60+45) + cos(60+45)

=[sin60cos45 +cos60sin45] + [cos60 cos45 - sin60sin45]

=cos 60 sin 45 +cos60 cos45

=1/2 * 1/sqrt2 + 1/2 * 1/sqrt2

=1/sqrt2

cos45°

Hope it helps u☺

hii kya hua gussa ho sorry :(