2024年5月22日 · I tried a question where you had to expand Cos (2x + x). I got Cos x but I just wanted to check with someone that the answer is right since there are no answers I can find to this question. cos2x.cosx-sin2x.sinx (cos2x=cos^2x -sin^2x) = (cos^2x-sin^2x)cosx-(2sin^2xcosx) = cos^3x-sin^2xcosx-2sin^2xcosx =cos^3x-3sin^2x.cosx ( sub 1-cos^2x=sin^2x ...

2024年5月15日 · As an experienced A-leveller, I can immediately recall cos ⁡ 2 x = 2 cos ⁡ 2 x − 1 = cos ⁡ 2 x − sin ⁡ 2 x = 1 − 2 sin ⁡ 2 x \cos 2x = 2 \cos^2 x - 1 = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x cos 2 x = 2 cos 2 x − 1 = cos 2 x − sin 2 x = 1 − 2 sin 2 x, but that's because I've used it so many times. Practice makes perfect.

2024年5月22日 · There are two ways that jump out at me immediately (although it is late, so forgive me if there is a quicker way): 1) Note that sin ⁡ 2 (2 x) = 1 2 [1 − cos ⁡ (4 x)] \sin ^2(2x) = \dfrac{1}{2}[1-\cos (4x)] sin 2 (2 x) = 2 1 [1 − cos (4 x)] so our integrand is now 1 2 cos ⁡ x − 1 2 cos ⁡ 4 x cos ⁡ x \dfrac{1}{2}\cos x - \dfrac{1 ...

1) Show that the equation tan 2x = 5 sinx 2x can be written in the form (1-5 cos 2x) sin 2x = 0 Here I just rearranged the first equation using: sin 2x / cos 2x = 5 sin 2x 2) Hence solve for 0 <= x <= 180 tan 2x = 5 sin 2x Answer to 1dp where appropriate. You must show clearly how answers are obtained I have attempted this in too many ways and ...

2021年12月6日 · sec^2= 1/cos^2x and then cos^2x = 1+ tan^2x. 0 Report. Reply 3. 3 years ago. username3563638. OP. 0. 16 ...

2024年6月20日 · If so, use the identity sin ⁡ 2 x + cos ⁡ 2 x ≡ 1 \sin^2 x + \cos^2 x \equiv 1 sin 2 x + cos 2 x ≡ 1 to put sin ⁡ 2 x \sin^2 x sin 2 x in terms of cos ⁡ x \cos x cos x (i.e. like in your first question), and then you'll get a quadratic in cos ⁡ x \cos x cos x.

2024年5月22日 · Yes. Set that equal to 1 and then solve for tan x, and find solutions as usual. EDIT: Although maybe it would be easier to solve tan(2x) = 1 for 2x between 0 and 4pi, and them half all of your solutions.

Can be written in (3sinx-1)² = 2 What I did: Sin²x Cos²x = 1 therefore cos²x = 1-sin²x 1-sin²x = 8sin²x-6sinx 1 = 9sin²x-6sinx Take out factor 3sin 1 = 3sin(3sin-2)... I didn't understand and got the question wrong, but I couldn't find any other solution for it

cos is not a number, it's like saying +^2. It doesn't make sense. It's only when you do cos of a number that you get another number. cosx is one number, so squaring = (cosx)^2. cos^2(x) is just a notation for that. Why do all that, it's obvious OP needs to be explained in simple terms.

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