Help needed with two problems (one a trig question, the other a world prob)

[DLxX]

1. Solve for theta in the interval 0<theta<2pi. Give answers in radians either correct to 3 decimal places or as exact values.

13cos^2 - 5cos = 12sin^2 - 6

3) Scientists use visual magnitude to measure the brighness of a star. The smaller the number, the brighter the star. An increase in magnitude of 1 unit represents a decrease of 60% in the brighness of a star.

a) The star Vega has a magnitude of 0. If a second star has a visual magnitude of -2.4, how many times brighter than Vega is this star?

b) If a star is 15% as bright as Vega is , what is its visual magnitude?

c)How is this scale different from the Richter scale?




For this next question all I need is a basic run through of what I have to do.

4) a and B are both angles in the second quadrant. If sin a = 1/3 and sin B = 2/3, find the exact value of cos (a +b).

Thanks

 

[HallsofIvy]

1. Solve for theta in the interval 0<theta<2pi. Give answers in radians either correct to 3 decimal places or as exact values.

13cos^2 - 5cos = 12sin^2 - 6

The first thing you should really try doing is copying the problem correctly! That is not the problem you were given because it doesn't make sense! I suspect that the problem is 13 cos²θ- 5 cosθ= 12 sin²θ- 6.
Replace sin²&theta with 1- cos²θ and you have a quadratic equation for cosθ solve for cosθ and then then use a calculator.

3) Scientists use visual magnitude to measure the brighness of a star. The smaller the number, the brighter the star. An increase in magnitude of 1 unit represents a decrease of 60% in the brighness of a star.

a) The star Vega has a magnitude of 0. If a second star has a visual magnitude of -2.4, how many times brighter than Vega is this star?

\
A "decrease" of 60% is the same as 40% or 0.4. -2.4 is a decrease in magnitude of 2.4 so an increase in brightness of 0.4 2.4 times: careful: that is not 2.4*0.4 but (0.4)^2.4.

b) If a star is 15% as bright as Vega is , what is its visual magnitude?

"15% as bright" is a decrease of 100-15= 85= (85/40)(40)= (2.125)(40) %

c)How is this scale different from the Richter scale?

Well, how is the Richter scale defined?

For this next question all I need is a basic run through of what I have to do.

4) a and B are both angles in the second quadrant. If sin a = 1/3 and sin B = 2/3, find the exact value of cos (a +b).

If sin a= 1/3 what is cos a? If sin B= 2/3 what is cos B? Do you know a formula for cos(a+b) in terms of cos(a), sin(a), cos(b), sin(b)?

Thanks

 

[tacman]

1. To solve for theta in the given interval, we can use the double angle formula for cosine: cos(2theta) = 2cos^2(theta) - 1. We can rewrite the given equation as 13cos^2(theta) - 5cos(theta) - 12sin^2(theta) + 6 = 0. Substituting the double angle formula, we get 13(2cos^2(theta) - 1) - 5cos(theta) - 12(1-cos^2(theta)) + 6 = 0. Simplifying, we get 26cos^2(theta) - 17cos(theta) - 6 = 0. Solving for cos(theta) using the quadratic formula, we get cos(theta) = 3/13 or -2/13. Since we are looking for values in the interval 0<theta<2pi, we can eliminate the negative value and get cos(theta) = 3/13. Taking the inverse cosine of this value, we get theta = 1.302 radians or approximately 0.913 radians to 3 decimal places.

2. a) To find how many times brighter the second star is compared to Vega, we can use the formula for magnitude difference: m1 - m2 = 2.5log(b2/b1), where m is the magnitude and b is the brightness. Plugging in the given values, we get 0 - (-2.4) = 2.5log(b2/1). Simplifying, we get log(b2) = 2.4/2.5 = 0.96. Taking the antilog, we get b2 = 9.5. This means that the second star is 9.5 times brighter than Vega.

b) To find the magnitude of a star that is 15% as bright as Vega, we can use the same formula and solve for m2: m1 - m2 = 2.5log(b2/b1), where m1 = 0 and b1 = 100%. Plugging in the values, we get 0 - m2 = 2.5log(15%/100%). Simplifying, we get m2 = -2.5log(0.15) = 2.5(-0.823) = -2.058. Therefore, the visual magnitude of this star