May 9th, 2019, 01:04 PM  #1 
Newbie Joined: Apr 2019 From: Malawi Posts: 19 Thanks: 0 Math Focus: Trigonometry and calculus  Trigonometric Quadrant
Suppose that A is a second quadrant angle with sin A = 1/4. Determine the exact value of sin(A/2). The problem has made me crazy. Help please Last edited by Khoxy; May 9th, 2019 at 01:11 PM. 
May 9th, 2019, 01:44 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716 
$\sin\frac{A}{2}=\sqrt{\frac{1\cos A}{2}}$ where $\cos A=\sqrt{\frac{15}{16}}$. The sign for $\cos A$ is negative, since $A$ is in the second quadrant. Ans.=0.984250984251476 Last edited by skipjack; May 9th, 2019 at 01:59 PM. 
May 9th, 2019, 02:02 PM  #3 
Newbie Joined: Apr 2019 From: Malawi Posts: 19 Thanks: 0 Math Focus: Trigonometry and calculus 
Mathman I don't understand how you come up with cos A. Would you clarify?

May 9th, 2019, 02:10 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
cos²(A) ≡ 1  sin²(A) and cos(A) ≡ 1  2sin²(A/2) $\sqrt{\small8 + 2\sqrt{15}}/\small4 = 0.9920$... 
May 10th, 2019, 12:49 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716 
$\sin{\frac{A}{2}}=0.992029696267167$ Arith. error in previous. $\cos^2A=1\sin^2A=1\frac{1}{16}=\frac{15}{16}$. Last edited by skipjack; May 10th, 2019 at 10:35 PM. 

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