Lecture Notes Trigonometric Identities 1 Sample Problems page 1 Prove each of the following identities. 1. tan x sin x + cos x = sec x 2. 1 1 + tan x = tan x sin x cos x sin x cos2 x = sin3 x + 1 + sin cos = 2 cos 8. 1 2 cos2 x = tan2 x 1 tan2 x + 1 cos x 1 sin x 2 cos2 x 9. sec x + tan x = 10. sin4 x 11. (sin x 3. sin x 4. cos 1 + sin cos4 x = 1 cos x)2 + (sin x + cos x)2 = 2 cos x 5. 1 sin x sin4 x 6. sin2 x 7. 1 cos x = 2 tan x 1 + sin x sin2 x + 4 sin x + 3 3 + sin x 12. = 2x cos 1 sin x 13. cos x 1 sin x tan x = sec x 1 + sin x cos2 x cos4 x =1 cos2 x cos x sin x = cos x 1 + sin x 14. tan2 x + 1 + tan x sec x = Practice Problems Prove each of the following identities. 1. tan x + 2. cos x 1 …show more content…
sin4 x cos4 x = 1 Solution: LHS = sin4 x cos4 x = sin2 x = 1 sin2 x cos2 x = 1 11. (sin x cos x)2 + (sin x + cos x)2 = 2 Solution: cos2 x = sin2 x + cos2 x sin2 x cos2 x cos2 x cos2 x = 1 2 cos2 x = RHS 2 LHS = (sin x cos x)2 + (sin x + cos x)2 = sin2 x + cos2 x 2 sin x cos x + sin2 x + cos2 x + 2 sin x cos x = 2 sin2 x + 2 cos2 x = 2 sin2 x + cos2 x = 2 1 = 2 = RHS c copyright Hidegkuti, Powell, 2009 Last revised: March 16, 2011 Lecture Notes Trigonometric Identities 1 page 4 12. sin2 x + 4 sin x + 3 3 + sin x = 2x cos 1 sin x Solution: LHS = sin2 x + 4 sin x + 3 (sin x + 1) (sin x + 3) (sin x + 1) (sin x + 3) = = = 2 cos2 x (1 + sin x) (1 sin x) 1 sin x sin x + 3 = = RHS 1 sin x 13. cos x 1 sin x Solution: tan x = sec x cos x cos x sin x cos2 x sin x (1 sin x) cos2 x sin x + sin2 x tan x = = = LHS = 1 sin x 1 sin x cos x cos x (1 sin x) cos x (1 sin x) 2 2 cos x + sin x sin x 1 sin x 1 = = = = RHS cos x (1 sin x) cos x (1 sin x) cos x 1 + sin x 14. tan2 x + 1 + tan x sec x = cos2 x Solution: sin2 x sin x 1 sin2 x cos2 x sin x LHS = tan x + 1 + tan x sec x = +1+ = + + 2x 2x 2x cos cos x cos x cos cos cos2 x 2 2 1 + sin x sin x + cos x + sin x = = RHS = 2x cos cos2 x 2 c copyright
Find the angle between the vectors u = (–5, 1, 3) and v = (–2, 1, –2).
The area of a right triangle is 2. The length of the base is the same as the length of the height. Find this length.
Find the number that 5 times itself is the same as 3 times that number plus 2.
Also I still do not quit understand how this amounts to 8 ------> (1+1)**(5-2) = 8 why 2 * ? and what is the value?
You must show all steps and provide any evidence needed in your solution to receive full credit.
Each angle in the corner of $ABCD$ is $90^{\circ}$\hfill ---\ since $ABCD$ is a square.\\
3 x 2 + -2 x 4 + 3 x 8 + -2 x 1= 6 -8 +24 -2= 20
We also assume that there are 20\% more girls than boys in a class. Assume that there are 70 learners in a class, then 20\% of 70 is 14, let $b$ be the number of boys, then there are $b+14$ girls in the class. Now $b + 14 + b = 70$ implies that $b = 28$ and there are 42 girls in the class. Therefore this means 60\% of the class are girls and 40\% are boys. I didn't give a reason for number 10) because I wanted to make some assumptions and state the problem correctly. This problem was giving me a stress because I could not figure it out. Number 8) was not clear for me because I could not understand if this 10\% we talking about is for the whole class or is for the number of boys. These are kinds of problems which will enable learners to understand the meaning of percentage and ratios. They are challenging and consuming time. It was not going to be easy to figure them out if I wasn't using the equations. In this activity, I learn much more on percentages and
sin2 (d2r) = sin2 ( Φ2 - Φ12) + cos (Φ1) cos (Φ2) sin2 ( λ2 - λ12)
+ (5% are 450 sec) + (3% are 125 sec) + (2% are 120 sec)
It is SOH, CAH, TOA. SOA is sin=Opposite/Hypotenuse, CAH is cos=Adjacent/Hypotenuse, and TOA is tan=Opposite/Adjacent. So Lets take an angle here so lets say that we are looking for angle B. So we will call that angle Theta. So lets figure out how to find the sin, cos, and tan, or Theta. We want to first focus on the Sin or Theta. So look up at the SOH, CAH, TOA and we see that sin is Opposite over Hypotenuse. So Sin of Theta = 3/5. Why so? Because the three is opposite of the angle Theta and the hypotenuse is across from the 90o angle. The Cos of Theta =4/5. And The Tan of Theta = ¾. So now that we have the basics down, we are now going to learn how to find the measurements. So we start off with the Pythagorean theorem so we can find out what the hypotenuse is. So h2=72 +42. h2=49+16. h2= 65. h=√65. So now lets find the trig function angle Theta. We are going to find the
Let that the sine formula give the number of hours of daylight in Adelaide when input any day of the year. Letting t be the day of two years (from 1 to 730), and T be the amount of daylight. The above figure 2 shows the graph of this equation.
||x||= √(1*1 + 1*1 + 0*0 + 1*1 + 0*0 + 1*1) = √4 =2
–x3 – x5 + x13 + x14 + x22 + x23 + x24 = 0
#17. What is the sum of the areas of the two triangles formed in number 16?