How do you solve these math problems?

[Kamen rider baron]

1.  Cos (x +pi/3)= 1/2

 

2. Show using trig identities that cos^4p-sin^4p = cos2p

 

use double angle formula to show that: cos^2p = 1/2 + 1/2cos2P

 

Solve for x, x is between 0 and 2pi

 

sin^2 X = 2-cos x

[Vioh]

1.  Cos (x +pi/3)= 1/2

 

2. Show using trig identities that cos^4p-sin^4p = cos2p

 

3. Use double angle formula to show that: cos^2p = 1/2 + 1/2cos2P

 

4. Solve for x, x is between 0 and 2pi --> sin^2 X = 2-cos x

 

For Question 1:

 

 

and where 'k' can be any integer

 

and --> These are the general solutions

 

If we have to solve for x, where x is between 0 and 2 pi, then 'k' can only have the values of 0 or 1. In that case, the solutions are: x = 0, x = 4pi/3, and x = 2pi

 

For Question 2 (LHS = Left-hand side, RHS = Right-hand side)

 

Transforming the LHS using simple algebra, we have:

 

Using pythagorean identity, in which , then

 

Now, transforming the RHS using double angle identity, we have:

 

We see that the LHS & RHS are equal, so we have shown that

 

For Question 3:

 

Transforming the RHS by using the double-angle identity, we have:'

 

Using pythagorean identity, in which , we have:

--> This is exactly the LHS of the equation that we have to prove!

 

For Question 4:

 

Applying the pythagorean identity on the LHS, we have:

 

The equation above is clearly a quadratic equation that doesn't have any real solutions (only imaginary solutions). And since x can only be real (as it's between 0 & 2pi), thus the equation doesn't have any solutions!

 

If you have any further questions, feel free to ask!

[harihrrnn]

 

1.  Cos (x +pi/3)= 1/2

 

2. Show using trig identities that cos^4p-sin^4p = cos2p

 

3. Use double angle formula to show that: cos^2p = 1/2 + 1/2cos2P

 

4. Solve for x, x is between 0 and 2pi --> sin^2 X = 2-cos x

 

For Question 1:

 

 

and where 'k' can be any integer

 

and --> These are the general solutions

 

If we have to solve for x, where x is between 0 and 2 pi, then 'k' can only have the values of 0 or 1. In that case, the solutions are: x = 0, x = 4pi/3, and x = 2pi

 

For Question 2 (LHS = Left-hand side, RHS = Right-hand side)

 

Transforming the LHS using simple algebra, we have:

 

Using pythagorean identity, in which , then

 

Now, transforming the RHS using double angle identity, we have:

 

We see that the LHS & RHS are equal, so we have shown that

 

For Question 3:

 

Transforming the RHS by using the double-angle identity, we have:'

 

Using pythagorean identity, in which , we have:

--> This is exactly the LHS of the equation that we have to prove!

 

For Question 4:

 

Applying the pythagorean identity on the LHS, we have:

 

The equation above is clearly a quadratic equation that doesn't have any real solutions (only imaginary solutions). And since x can only be real (as it's between 0 & 2pi), thus the equation doesn't have any solutions!

 

If you have any further questions, feel free to ask!

 

wow,,. you are pretty good at math..!