2015年12月17日 · How do you prove cos 2x = cos2 x − sin2 using other trigonometric identities?

How do you find the exact values of sin 2u, cos 2u, and tan 2u using the double-angle formulas given tanu = 3 4, 0 <u <π 2?

2018年5月4日 · (sinx −cosx)2 + (sinx + cosx)2 = 2 sin2x −2sinxcosx +cos2x + sin2x +2sinxcosx +cos2x = 2 red terms equal 1 from the Pythagorean theorem also, blue terms equal 1 So 1−2sinxcosx + 1+2sinxcosx = 2 green terms together equal 0 So now you have 1 …

2015年6月5日 · Use the trig identity cos (a + b) cos(a + a) = cosa.cosa − sina.sina = cos2a − sin2a

2017年2月13日 · The formula for a circle centred at the origin is x2 +y2 = r2 That is, the distance from the origin to any point (x,y) on the circle is the radius r of the circle. Picture a circle of radius r centred at the origin, and pick a point (x,y) on the circle: graph { (x^2+y^2-1) ( (x-sqrt (3)/2)^2+ (y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]} If we draw a line from that point to the origin, its ...

2018年6月12日 · Remember the equation cos2x + sin2x = 1? Well the x refers to any number so if your number is 2x, then cos22x + sin22x = 1 You can also prove this by using the double angle formula cos2(2x) +sin2(2x) = (cos2x −sin2x)2 +(2sinxcosx)2 = cos4x − 2sin2xcos2x + sin4x +4sin2xcos2x = cos4x + 2sin2xcos2x + sin4x = (cos2x +sin2x)2 = 12 = 1

2018年1月31日 · Prove: 1- cos (x) = 2sin^2 (x/2) Digress, for a moment, to the identity cos (2u) = 1 - 2sin^2 (u) Substitute u = x/2: cos (x) = 1-2sin^2 (x/2) Add 2sin (x/2)-cos (x) to both sides: 2sin^2 (x/2) = 1-cos (x) Return to the original equation and make the above substitution: 1- cos (x) = 1-cos (x) Q.E.D.

The functions sine, cosine and tangent of an angle are sometimes referred to as the primary or basic trigonometric functions. The remaining trigonometric functions secant (sec), cosecant (csc), and cotangent (cot) are defined as the reciprocal functions of cosine, sine, and tangent, respectively. Trigonometric identities are equations involving the …

2015年5月23日 · It depends where you want to start. If you know the Taylor series for e^z, sin theta and cos theta, together with the basic properties of i = sqrt(-1), then you can easily find that: e^(itheta) = cos theta + i sin theta Then: cos 2theta + isin 2theta = e^(2itheta) = (e^(itheta))^2 = (cos theta + isin theta)^2 = cos^2theta + 2i cos theta …

2015年4月22日 · Recall the Pythagorean Identity sin2x +cos2x = 1 Which can be manipulated into this form: cos2x = 1 − sin2x In our equation, we can replace cos2x with this to get 1 − sin2x −sin2x, which simplifies to 1 − 2sin2x. We have just verified the identity ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ 2 2cos2x − ...