Given that cos A = √3 - 1, find the value of sin^2A in the form p √3 + q

Trigonometric identities
in Trigonometry Answers by Level 1 User (240 points)

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Best answer

cos2(A)=(√3-1)2=3-2√3+1=4-2√3. sin2(A)=1-cos2(A)=1-(4-2√3)=2√3-3 (p=2, q=-3).

by Top Rated User (1.2m points)
selected by

Related questions

1 answer
sec/sin - sin/cos -= cot x
asked Mar 4, 2013 in Trigonometry Answers by anonymous | 1.5k views
1 answer
If sinѳ = ½(x+1/x) , show that sin 3ѳ+1/2(x^3+1/x^3) = 0
asked Aug 8, 2013 in Trigonometry Answers by Barbie Girl | 993 views
1 answer
Find P(X,Y), given P(x,y).
asked Apr 28, 2015 in Geometry Answers by Rod Top Rated User (1.2m points) | 1.3k views
1 answer
1 - sec x cos x = tan x cot x - 1
asked Jan 11, 2013 in Trigonometry Answers by anonymous | 2.9k views
1 answer
Sin^2+Sin0=0 for 0 is greater or equal to 2 x 3.14
asked Mar 25, 2014 in Trigonometry Answers by anonymous | 663 views
1 answer
given the function: f(x)=(x^(2))/(x+4)
asked Aug 5, 2013 in Trigonometry Answers by Lalala | 713 views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!

Most popular questions within the last 3000 days

  1. 50+50-25X0+2+2= (19)
  2. it takes 10 road workers 5 days to complete a road repair when working 2 hours a day.Working at the same pace,how many days will it take 2 people working 5 hours a day to finish it? (3)
  3. Solve 15 -1(12÷4+1)? (4)
  4. 8:4(5+2) (3)
  5. How can i represent 1/3; 3/3; 7/3; 10/3 and 15/3 on a number line? (1)

Most popular tags

87,516 questions
100,285 answers
2,420 comments
736,082 users