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Write 9sin(x) + 12 cos(x) in the form Rsin(x+y) and hence solve 9sin(x) + 12 cos(x) = 3

9sin(x) + 12 cos(x) = Rsin(x+y) =R(sin(x)cos(y)+cos(x)sin(y))= (Rcos(y))sin(x) + (Rsin(y))cos(x)Therefore by matching the coefficientsRsin(y)=12, Rcos(y)=9 [1]SoRsin(y)/Rcos(y) = 12/9 = 4/3, therefore tan(y)...
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Answered by James H. • Maths tutor
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g(x) = e^(x-1) + x - 6 Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, where x<6

0 = e x-1 + x - 6 e x-1 = 6-x x-1 = ln (6-x) -&gt; here we have taken the natural log of both sides, but it only shows on one side as the natural log of e is 1.x = ln (6-x) + 1Question taken from Edexcel 201...
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Answered by Sumrah N. • Maths tutor
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Find the coefficient of x^4 in the expansion of: x(2x^2 - 3x + 1)(3x^2 + x - 4)

Only the terms which will form the x 4 term need to be considered:x * 2x 2 * x + x * -3x * 3x 2 = -7x 4 Therefore the answer is -7.
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Answered by Finn H. • Maths tutor
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An object of mass 3kg is held at rest on a rough plane. The plane is inclined at 30º to the horizontal and has a coefficient of friction of 0.2. The object is released, what acceleration does the object move with?

We need to use Newtons law F=ma going down the slope. We can see that the only forces acting in this direction are the component of the weight and friction, so we have that: F = Wsin30 - μR = 3a We have that...
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Answered by Asha D. • Maths tutor
3907 Views

Show that (sec(x))^2 /(sec(x)+1)(sec(x)-1) can be written as (cosec(x))^2.

( sec 2 (x))/((sec(x)+1)(sec(x)-1))Then, by the rule of 'difference of two squares', we know that this equals= (sec 2 (x))/(sec 2 (x)-1)= (sec 2 x/tan 2 x)since 1+tan 2 (x)=sec 2 (x), we get sec 2 (x)-1=tan ...
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Answered by Rishi S. • Maths tutor
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