Show that cos (θ-40) = sin (θ+50). Hence, or otherwise, solve the equation cos(θ-40) = sin(θ+50), giving all solutions between 0° and 360°.
Answer
To show that cos(θ-40) is equal to sin(θ+50), we can use the identities:
cos(a-b) = cos(a)cos(b) + sin(a)sin(b)
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
Let's apply these identities:
cos(θ-40) = cos(θ)cos(40) + sin(θ)sin(40) [Using the first identity]
sin(θ+50) = sin(θ)cos(50) + cos(θ)sin(50) [Using the second identity]
Now we need to show that these two expressions are equal:
cos(θ)cos(40) + sin(θ)sin(40) = sin(θ)cos(50) + cos(θ)sin(50)
Now, we can rearrange the terms to better compare them:
cos(40)cos(θ) + sin(40)sin(θ) = cos(50)sin(θ) + sin(50)cos(θ)
By observing the pattern of the right side, we can rearrange it using the commutative property of addition:
cos(40)cos(θ) + sin(40)sin(θ) = sin(θ)cos(50) + cos(θ)sin(50)
cos(40)cos(θ) + sin(40)sin(θ) = sin(θ)cos(50) + sin(50)cos(θ)
Now, we can match the terms on both sides and see that they are indeed equal:
cos(40)cos(θ) + sin(40)sin(θ) = sin(θ)cos(50) + sin(50)cos(θ)
Thus, we have shown that cos(θ-40) = sin(θ+50).
To solve the equation cos(θ-40) = sin(θ+50), we can use the double angle formula for sine:
sin(2θ) = 2sin(θ)cos(θ)
We can rewrite the equation as follows:
cos(θ-40) = sin(θ+50)
cos(θ-40) = cos(90 - (θ+50)) [Using the identity sin(θ) = cos(90-θ)]
θ - 40 = 90 - (θ + 50) [Since cos(x) = cos(-x) for all x]
θ - 40 = 90 - θ - 50
2θ = 90 - 40 - 50
2θ = 0
θ = 0
Therefore, the only solution between 0° and 360° is θ = 0.
To prove the equation cos(θ-40) = sin(θ+50), we can use the trigonometric identities and basic algebraic manipulation.
Let's start by using the difference of angles identity for cosine:
cos(A - B) = cos(A)cos(B) + sin(A)sin(B)
In this case, A = θ and B = 40, so we have:
cos(θ - 40) = cos(θ)cos(40) + sin(θ)sin(40)
Now, let's use the sum of angles identity for sine:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
Using this identity with A = θ and B = 50, we get:
sin(θ + 50) = sin(θ)cos(50) + cos(θ)sin(50)
Comparing this result with the earlier equation, we can see that the right side of both equations has the same terms but in different orders. Therefore, we can rewrite the second equation as:
sin(θ + 50) = cos(θ)sin(40) + sin(θ)cos(40)
Since the right sides of both equations are now identical, we can conclude that:
cos(θ - 40) = sin(θ + 50)
Now let's solve the equation cos(θ - 40) = sin(θ + 50) to find the solutions between 0° and 360°.
cos(θ - 40) = sin(θ + 50)
Let's rearrange the equation:
cos(θ - 40) - sin(θ + 50) = 0
Using the sum-to-product identity for cosine, we can rewrite cos(θ - 40) as:
cos(θ - 40) = cos(θ)cos(40) + sin(θ)sin(40)
Now substitute this into the equation:
cos(θ)cos(40) + sin(θ)sin(40) - sin(θ + 50) = 0
Now let's simplify the equation using trigonometric identities:
cos(θ)cos(40) + sin(θ)sin(40) - [sin(θ)cos(50) + cos(θ)sin(50)] = 0
cos(θ)cos(40) - sin(θ)cos(50) + sin(θ)sin(40) - cos(θ)sin(50) = 0
Rearranging the terms:
cos(θ)(cos(40) + sin(50)) - sin(θ)(cos(50) - sin(40)) = 0
Now we have two factors, one with cos(θ) and the other with sin(θ):
cos(θ)(cos(40) + sin(50)) - sin(θ)(cos(50) - sin(40)) = 0
Setting each factor equal to zero:
cos(θ) = 0 or cos(40) + sin(50) = 0
sin(θ) = 0 or cos(50) - sin(40) = 0
The solutions for cos(θ) = 0 are θ = 90° and θ = 270°.
To find the solutions for cos(40) + sin(50) = 0, we need to use special angle values. Let's solve it:
cos(40) + sin(50) = 0
cos(40) = -sin(50)
Using the fact that sin(𝜋/2 - A) = cos(A), we can rewrite the equation:
-sin(𝜋/2 - 40) = -sin(50)
This implies that:
𝜋/2 - 40 = 50
Simplifying:
𝜋/2 = 90
Therefore, there are no solutions for cos(40) + sin(50) = 0 within the range of 0° to 360°.
In summary, the solutions to the equation cos(𝜃 - 40) = sin(𝜃 + 50) between 0° and 360° are 𝜃 = 90° and 𝜃 = 270°.
by definition, costheta=sin(90-theta)
let theta=θ-40
then cos(θ-40)=sin(90-θ+40)
but sin (90-x)=sin(90+x)
then cos(θ-40)=sin(90+θ-40)=sin(θ+50)