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...equal 3^b, so b = 1 That's the only way to get 8 as a factor of the

**LHS**and 3 as a factor of the RHS. This is the same way everyone...4 Answers · Science & Mathematics · 22/01/2018

To prove : cosx + cos2x + cos3x = cos2x(1 + cos2x)

**LHS**= cosx + cos2x + ( cos(2x + x) = cosx + cos2x + cos2x* cosx - sin2x*sinx = cosx + cos2x + cos2x...2 Answers · Science & Mathematics · 02/01/2018

... = (a + b)(a² - ab + b²) Let a = sinx, b = cosx : ∴

**LHS**= sin³x + cos³x = a³ + b³ = (a + b)(a² - ab...2 Answers · Science & Mathematics · 11/12/2017

... = 100 N But note that the two forces on the

**LHS**are parts of two different action/reaction pairs. As for...2 Answers · Science & Mathematics · 08/12/2017

RHS = (tanx-1)(cotx+1) = 1+tanx-1/tanx-1 = tanx - 1/tanx = tanx - cotx =

**LHS**4 Answers · Science & Mathematics · 06/12/2017

.../m * 1.5m * (0.5m + 2.5m) = 27 kN·m The moment due to the

**LHS**of the load is 6kN/m * 1.5m * 0.5m = 4.5 kN·m and so ...1 Answers · Science & Mathematics · 10/11/2017

...from both sides -4 < -2x step 3 rewrite with the -2x term on the

**LHS**-2x > -4 Which shows if you multiply the original inequality by -1...2 Answers · Science & Mathematics · 04/11/2017

...500 let u = tanθ du/dθ = sec²θ let v = x/500 dv/dx = 1/500 differentiate

**LHS**WRT t using the chain rule du/dt = du/dθ dθ/dt = sec²θ...5 Answers · Science & Mathematics · 09/04/2017

**LHS**= (1+cosx)/sinx = [ 1 + (2cos^2(x/2) - 1)]/2sin(x/2)cos(x/2) = 2cos^2(x/2)/2sin(x/2)cos(x/2) = cos(x/2)/cos(x/2) * cos(x/2)/sin(x/2) = 1 * cos(x/2)/sin(x/2) = cot(x/2) = RHS2 Answers · Science & Mathematics · 30/11/2016

...there exist x,y > 0 such that (x+y) (1/x + 1/y) < 4 simplify the

**LHS**and we get (x+y)^2/(xy) < 4 (x+y)^2/(xy) - 4 < 0...2 Answers · Science & Mathematics · 02/11/2016