2015年7月21日 · I found: sin (15°)=0.258 Using the Half Angle Formula: color (red) (sin^2 (x)=1/2 [1-cos (2x)]) with x=15° and 2x=30° you get: sin^2 (15°)=1/2 [1-cos (30 ...

2016年12月23日 · How do you find the value of sin 15 using the double or half angle formula?

2015年9月28日 · Find sin (15) Ans: sqrt (2 - sqrt3)/2 Use the trig identity: cos 2a = 1 - 2sin^2 a cos (30) = sqrt3/2 = 1 - 2sin^2 (15) 2sin^2 (15) = 1 - sqrt3/2 = (2 - sqrt3)/2 sin^2 (15) = (2 - sqrt3)/4 sin (15) = +- sqrt (2 - sqrt3)/2. Since arc (15) deg is in Quadrant I, its sin is positive. Then, sin (15) = sqrt (2 - sqrt3)/2. Check by calculator. sin 15 = 0.26 sqrt (2 + sqrt3)/2 = 0.52/2 = 0.26. OK

2016年9月29日 · sin 165^@ = 1/4 (sqrt (6)-sqrt (2)) Some things we will use: sin (theta) = sin (180^@ - theta) sin (alpha-beta) = sin alpha cos beta - sin beta cos alpha sin 45^@ = cos 45^@ = sqrt (2)/2 sin 30^@ = 1/2 cos 30^@ = sqrt (3)/2 Hence we find: sin 165^@ = sin (180^@-165^@) color (white) (sin 165^@) = sin 15^@ color (white) (sin 165^@) = sin (45 ...

mathisfun did a nice job of showing you how to reconcile your answers. Another way of showing that sin(45 ∘) + sin(15 ∘) = sin(75 ∘) is to use the formula sinα + sinβ = 2sin(α + β 2)cos(α − β 2)

2018年3月20日 · sin45cos15 +cos45sin15 It is in the form sinacosb + cosasinb But we know sin(a +b) = sinacosb +cosasinb ∴ sin45cos15 +cos45sin15 = sin(45 +15) = sin60 = √3 2

I am having problems understanding how to solve 14(6–√ − 2–√) 1 4 (6 − 2), find the exact value of sin15∘ sin 15 ∘. I have the answer, but I need help understanding the methods to achieve the answer. Thank you.

I rewrote the statement as $$ \\sin(30^° + 15^°) + \\sin(15^°) = \\cos(15^°). $$ Then I got $$ (\\sqrt{3}-2) \\sin(15^°) = \\cos(15^°). $$

2015年3月27日 · The answer I got when trying to solve it was $\\sqrt{\\frac{1 - \\sqrt3}{2} }$ but the book says it's $\\sqrt{ \\frac{2 - \\sqrt3}{2}}$ and I don't know how the two on the top half gets there.

2023年2月21日 · Here is another way that gives sin 15 and sin 75 pretty quickly. (Assuming we know how to use geometry to get sin/cos 30/45.) We define isosceles ABC with vertex A of 30 degrees. We draw in BD so BD=BC, and we draw DE perpendicular to AB. Let x = sin 15. Then BC=2x. BD=BC. Since BDE is an isosceles right triangle, BE = DE = x√2. But ADE is a 30-60-90 triangle, so AE = x√6. That makes sin15 ...