Solve the following questions
(1) SinB=Cos35 degrees
(2) Cos286 degrees
(3) SinB=Cos3B
(4) Cos2B=Sin3B
(5) Cos y= Sin(y+22 degrees)
(1) SinB=Cos35 degrees
Draw the triangle. COsine = SINE of COmplement: 55°
(2) Cos286 degrees
= cos(270+16) = cos270cos16 - sin270sin16 = sin16°
(3) SinB=Cos3B
sinB-cos3B = 0
sin(π/2-B)-cos(3B)=0
now using the sum-to product formulas, that gives
2sin(B-π/4)sin(2B-π/4) = 0
So, since sin(0) = sin(π) = 0, just solve for B
(4) Cos2B=Sin3B
same way as above
(5) Cos y= Sin(y+22 degrees)
90-y = y+22
y = 34°
Sure! Let's solve each of the questions step by step.
(1) SinB = Cos35 degrees:
To solve this equation, we can use the trigonometric identity: Sin(B) = Cos(90 degrees - B).
In this case, B = 90 degrees - 35 degrees = 55 degrees.
So, the solution is B = 55 degrees.
(2) Cos286 degrees:
To find the value of Cos286 degrees, we can use the unit circle or a calculator.
If we use the unit circle, we can see that the reference angle for 286 degrees is 74 degrees (360 degrees - 286 degrees).
Cosine is positive in the fourth quadrant, so the value of Cos286 degrees is the same as the value of Cos74 degrees, which is approximately 0.279.
So, Cos286 degrees ≈ 0.279.
(3) SinB = Cos3B:
To solve this equation, we can use the trigonometric identity: Sin(B) = Cos(90 degrees - B) and simplify.
Sin(B) = Sin(90 degrees - 3B)
Since Sin and Cos are equal only when their angles are complementary, we can set 90 degrees - B = 90 degrees - 3B.
Simplifying the equation, we get B = 30 degrees.
(4) Cos2B = Sin3B:
To solve this equation, we can use the trigonometric identity: Cos(2B) = 1 - 2Sin^2(B) and Sin(3B) = 3Sin(B) - 4Sin^3(B).
We can substitute these identities into the equation and simplify:
1 - 2Sin^2(B) = 3Sin(B) - 4Sin^3(B)
Rearranging the equation, we have 4Sin^3(B) - 2Sin^2(B) - 3Sin(B) + 1 = 0
This equation can be factored as (2Sin(B) - 1)(2Sin^2(B) - 2Sin(B) -1) = 0
The solutions for Sin(B) are either 1/2 or the solutions of 2Sin^2(B) - 2Sin(B) - 1 = 0.
If Sin(B) = 1/2, then B = 30 degrees or B = 150 degrees.
For 2Sin^2(B) - 2Sin(B) - 1 = 0, we can use the quadratic formula to find the solutions, which are approximately B ≈ 9.189 degrees and B ≈ 170.811 degrees.
So, there are four solutions for B: B = 30 degrees, 150 degrees, 9.189 degrees, and 170.811 degrees.
(5) Cos(y) = Sin(y + 22 degrees):
To solve this equation, we can use the trigonometric identity: Sin(A + B) = Sin(A)Cos(B) + Cos(A)Sin(B).
Let's apply this identity to the equation:
Cos(y) = Sin(y)Cos(22 degrees) + Cos(y)Sin(22 degrees)
Rearranging the equation, we have:
Cos(y) - Cos(y)Sin(22 degrees) = Sin(y)Cos(22 degrees)
Factoring out Cos(y), we get:
Cos(y)(1 - Sin(22 degrees)) = Sin(y)Cos(22 degrees)
Dividing both sides by (1 - Sin(22 degrees)), we get:
Cos(y) = Sin(y)Cos(22 degrees) / (1 - Sin(22 degrees))
To solve for y, we need to take the arccosine of both sides. However, the equation involves both Sin(y) and Cos(y), so it cannot be easily simplified into a single function. Therefore, to find the numeric solution, we can either use numerical methods or a graphing calculator to find the intersection point.
I hope this helps! Let me know if you have any further questions.