Verify the following identity.
1) sin(x+y)
__________ = tanx+tany
cosxcosy
2) Write cos3x in terms of cosx
I cannot figure out how to solve for identities they are so hard. Any help appreciated!:)
Sure! I can help you verify the given identity and write cos3x in terms of cosx. Let's start with the first one:
1) To verify the identity sin(x+y) / (cosx * cosy) = tanx + tany, we can simplify both sides of the equation.
On the left-hand side (LHS):
sin(x+y) / (cosx * cosy)
Using the trigonometric identity sin(a+b) = sin(a)cos(b) + cos(a)sin(b), we can rewrite the numerator of the LHS as:
sin(x)cos(y) + cos(x)sin(y)
Dividing this by (cosx * cosy), we get:
(sin(x)cos(y) + cos(x)sin(y)) / (cosx * cosy)
Now, let's simplify the denominator:
cosx * cosy = cos(x)cos(y)
Dividing the numerator by the denominator, we have:
(sin(x)cos(y) + cos(x)sin(y)) / (cos(x)cos(y))
Using the trigonometric identity sin(a) / cos(a) = tan(a), we can rewrite the numerator as:
tan(x) + tan(y)
Therefore, the left-hand side (LHS) becomes:
tan(x) + tan(y)
On the right-hand side (RHS):
tanx + tany
Since the LHS and RHS are the same, we have verified the given identity sin(x+y) / (cosx * cosy) = tanx + tany.
Now, let's move on to the second question:
2) To write cos3x in terms of cosx, we can use the trigonometric identity cos(3a) = cos^3(a) - 3cos(a)sin^2(a).
In this case, let a = x. Therefore, we have:
cos(3x) = cos^3(x) - 3cos(x)sin^2(x)
Using the trigonometric identity sin^2(a) = 1 - cos^2(a), we can substitute it into the equation:
cos(3x) = cos^3(x) - 3cos(x)(1 - cos^2(x))
Expanding further, we get:
cos(3x) = cos^3(x) - 3cos(x) + 3cos^3(x)
Combining like terms, we have:
cos(3x) = 4cos^3(x) - 3cos(x)
Therefore, cos3x can be written in terms of cosx as 4cos^3(x) - 3cos(x).
I hope this helps you! Let me know if you have any further questions.
Sure! I'd be happy to help you with these two identities.
1) To verify the identity sin(x+y)/(cosx * cosy) = tanx + tany.
To prove this identity, we can start by using the trigonometric addition formula for sine:
sin(x+y) = sin(x)cos(y) + cos(x)sin(y).
Now, let's substitute this in the given identity:
sin(x+y) / (cosx * cosy) = (sin(x)cos(y) + cos(x)sin(y)) / (cosx * cosy).
Next, we can simplify the numerator:
= (sin(x)cos(y))/(cosx * cosy) + (cos(x)sin(y))/(cosx * cosy).
Now, we can rewrite each fraction using the quotient formula for tangent:
= tanx * coty + tany * cotx.
Finally, we can factor out a common cotangent factor:
= coty(tanx) + cotx(tany) = tanx + tany.
Hence, we have verified the given identity.
2) To write cos3x in terms of cosx.
We can expand cos(3x) using the trigonometric triple angle formula:
cos(3x) = cos(2x + x).
Applying the double angle formula to the first term:
cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x).
Using the double angle formulas for cosine and sine:
= (cos^2(x) - sin^2(x)) * cos(x) - (2sin(x)cos(x)) * sin(x).
Expanding further:
= cos^3(x) - sin^2(x) * cos(x) - 2sin^2(x) * cos(x).
Now, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1:
= cos^3(x) - (1 - cos^2(x)) * cos(x) - 2(1 - cos^2(x)) * cos(x).
Simplifying this expression:
= cos^3(x) - cos(x) + cos^3(x) - 2cos(x) + 2cos^3(x).
Combining like terms:
= 4cos^3(x) - 3cos(x).
Therefore, cos3x can be written as 4cos^3(x) - 3cos(x).
I hope this explanation helps you understand how to solve trigonometric identities! Let me know if you have any further questions.