cos 68° cos 8° + sin 68° sin 8° = ?
Cos68°cos8°+sin68°sin8°
We know,
cosAcosB+sinAsinB=cos(A-B)
Cos(68-8)
Cos60°
=1/2
cos130° cos40°+ sin130° sin40° is
cos(a-b) = Cos a Cos b + Sin a Sin b
Cos(130-40)°=cos90°=0
Which of the following is correct:
sin(x+y) = sinxcosy + cosxsiny
It is the formula
The value of Cos75°is equal to
Sin A = 1/√10 , Sin B = 1/√5 If A and B are both acute angles, then, A+B =?
Solution:- We know that,
sin theta = Opposite/Hypotenuse
therefore, SinA = 1/root10
CosA= 3/root10
similarly, SinB = 1/root5
CosB= 2/root5
now, multiply cos in A+B
Cs(A+B)= CosA x CosB - SinA x SinB substituting the value in above equation we get,
↔3/root10 x 2/root5 - 1/root 10 x 1/root 5
I.e 6/root50 - 1/root50
6-1/5root2. ........(root50=5root2)
5/5root21/ root 2
we know that, sin 45 = 1/ root 2 therefore sin theta/ Cos theta= 45
The value of sin 120° cos 60° + cos 120° sin 60° is equal to
Sin A Cos B + Cos A SinB =Sin( A + B)
= Sin(120+60)° = Sin (180)° = 0
The values of sin 105° is equal to
ANSWER :- a
Solution :- sin (105)° = sin (15 + 90) = cos 15°.
First find (cos 15°). Call cos 15° = cos x
Apply the trig identity:
cos2x = 2cos2 x−1
cos2 x = cos (30)° = √3/2 = 2cos2 x−1
2cos2 x = 1 + sqrt3/2 = (2 + sqrt3)/2
cos2 x = (2 + sqrt3)/4
cos x = cos 15° = (sqrt(2 + sqrt3)/2. (since cos 15° is positive)sin(105)° = cos(15)°
Chose which of the following expressions equals sinA + cosA.
SinA+cosA=√2(sinA/√2+cosA/√2)
=√2(sinAcosπ/4+cosAsinπ/4)
=√2sin(A+π/4)
by using sin(A+B) formula
The value of cos45°cos15° + sin 45° sin15° is
sin(60° + A) cos(30° – B) + cos(60° + A) sin(30° – B) is equal to
Sin(60+a)cos(30-b)+ cos(60+a)sin(30-b) =sin(60+a+30-b) sin(a)cos(b) + cos(a)sin(b)= sin(a+b) =sin(90+(a-b)) =cos(a-b)
sin(45° – A) cos(45° – B) + cos(45° – A) sin(45° – B)
It is equal to SinACosB + CosASinB = cos(A+B)
hence,
= Sin(45-A+45-B)
= Sin(90-(A+B))
= Cos(A+B)
There's a formula sinAcosB + cosAsinB = sin(A+B) so, here in this question, assume (45-A) as A and (45-B) as B so you'll get sin( 90 -(A+B)) which is cos (A+B)
Sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A =
Sin(n+1)ASin(n+2)A + Cos(n+1)ACos(n+2)A =Cos (n+1)ACos(n+2)A + Sin(n+1)ASin(n+2)A =Cos{A(n+2-n-1)} =cos (A.1) =cos A
cos(π/4 - x) cos (π/4 - y) -sin(π/4 - x) sin(π/4 - y) =
Cos(π/4-x)cos (π/4-y) - sin (π/4-x) sin(π/4-y)
= CosA*Cos B - Sin A*Sin B
=Cos (A+B)
=cos(π/4-x+π/4-y)
=cos(π/2-x-y)
=cos{π/2 - (x+y)}
= sin(x+y)
cos130° cos40° + sin130° sin40° is equal to
cos130°cos40°+sin130°sin40° = cos (130-40)° = cos90° =0
Which of the following is correct:
(a) cos (A+B)=cos A cos B - sin A sin B, a is incorrect
(d) cos(A-B)=cos A cos B + sin A sin B. {it is a formula)
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