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If f(x) = 4 sin x / 1 + cos x then find f'(x). Given f'(x) = 2 cos x - 3 sin x and f(0) = 6, find f(x). Given f(x) = 2x^3 + 4 \sin x + 3 \cos x , find (f^{-1})'(3) . Given f'(x) = 6 \cos x - 10 \sin x and f(0) = 3, \ fin...
Answer to: Write the following in terms of sin theta and cos theta, then simplify if possible. \csc\theta - \cot\theta\cos\theta By signing up,...
Step by step video, text & image solution for If alpha is a root of 25"cos"^(2) theta+ 5"cos" theta-12 = 0, (pi)/(2) lt alpha lt pi, " then sin"2 alpha is equal to by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.Updated on:21/07/...
Answer to: If cos theta = 1/3 and theta is in the 4 quadrant, Find sin(theta + pi/3) By signing up, you'll get thousands of step-by-step...
Explore trigonometric ratios? Learn how to do the trigonometric ratios sin, cos and tan. Understand the concept of similar triangles ratio in right triangle trigonometry. Related to this Question Explore our homework questions and answers library ...
2.1.438 Part 1 Section 17.15.2.19, frameset (Root Frameset Definition) 2.1.439 Part 1 Section 17.15.2.21, left (Left Border for HTML div) 2.1.440 Part 1 Section 17.15.2.23, longDesc (Frame Long Description) 2.1.441 Part 1 Section 17.15.2.30, name (Frame Name) 2.1.442 Par...
It was experimentally demonstrated that the BDS-2 satellite PIFCB showed a periodic variation and with some of the satellites corrected by PIFCB, the mean root mean square error (RMS) value of the GFIF phase combination was 5 mm, which was an improvement of 50%. Gong et al. [20] and ...
- } - if(A < 1e-4) discard; - outColor = vec4(color, A); - return; - } - - if(type == Airbrush){ - float tanTheta = sqrt(1.0 - cosTheta*cosTheta)/cosTheta; - float mid = pLocal.x - abs(pLocal.y)/tanTheta; - float A = alpha; - float transparency0 = d0 > r0...
[0 0 0] [1 0 0]] """print(csr) """ 只显示非零的数据和位置:1行2列的位置为2,3行1列的位置为1 (0, 1) 2 (2, 0) 1 """ from__future__importunicode_literalsimportnumpy as np a= np.arange(1,3)print(a, a.shape, sep='')#[1 2] (2,)b= np.array([{1,2,3}, ...