Question

please solve these questions:
1)sin (arc cos3/5)

2) sin cos(π/9)
​

Answer 1

Answer:

Let arccos(3 5) = A, then cosA = 3 5 and sinA = √1 − cos2A = √1 −(3 5)2 = √1 − 9 25 = 4 5 an as A = arccos(3 5) sin(arccos(3 5)) = 4 5

Step-by-step explanation:

Answer 2

Answer:

Given

cos

−1

(cos(

3

5π

))+sin

−1

(sin(

3

5π

))

We know, cosx=(sin

2

π

−x)

=cos

−1

(sin(

2

π

−

3

5π

))+sin

−1

(sin(

3

5π

))

=cos

−1

(sin(−

6

7π

))+sin

−1

(sin(

3

5π

))

=cos

−1

(−sin(

6

7π

))+sin

−1

(sin(

3

5π

))

Now, sin

−1

(sin

3

5π

)=−

3

π

=cos

−1

(−sin(

6

7π

))+(−

3

π

)

Now, cos

−1

(−sin

3

7π

)=

3

π

=(

3

π

)+(−

3

π

)

=0