+9673 1) [Logarithms]
Simplify \(\log_3(81\sqrt[3]{\dfrac{\sqrt{3+\ln1}}{3}})\)
2. [Complex numbers]
Find the value for \(\ln(\sqrt[n]{-1})\) in terms of n.
3. [Trigonometry]
(a) Express \(4\sin x + 3\cos x\) in the form of \(a\sin(x+\alpha)\)
(b) Solve the equation \(4\sin x + 3 \cos x = 2\) for the range \(0^{\circ} \leq x <2\pi\)
(x and alpha are in radians)
4. [Algebra]
(a) Factorize \(x^4 + x^2 + 1\)
(b) If \(y = x\sqrt x\), express \(x^2+x+1\) in terms of y(That means no 'x's in the final answer.)
(Please show steps.)
5. [Integral Calculus]
Compute \(\displaystyle \int\sqrt{\tan x}dx\)
(Do not show steps. This will make the whole forum lag.)
6. [Algebra]
(a) State the exact value of the golden ratio.
(b) Find a quadratic equation whose one of the roots is the golden ratio. The equation must be in the simplest form.
7. [Algebra]
If 3xy = a + 2, find the value of \((27^x)^y\), in the form of \(R\cdot Q^a\), where Q and R are positive integers.
(So easy this one.......)
8. [Differential Calculus]
Differentiate \(u(x)\), where \(u(x)\) is the unit-step function.
9. [Complex numbers]
Compute \(\sqrt{1+i}\)
10. [Trigonometry]
Simplify \(\cos \arctan x\).
+5265 x4+x2+1
(x2-x+1)(x2+x+1)
I don't believe that there're many steps to this one, but I took a shot at it. :/
1) 4/3^(1/6)
2) ((i π)/2)^n
4a) (x^2 - x + 1) (x^2 + x + 1)
5) integral sqrt(tan(x)) dx = (-2 tan^(-1)(1 - sqrt(2) sqrt(tan(x))) + 2 tan^(-1)(sqrt(2) sqrt(tan(x)) + 1) + log(tan(x) - sqrt(2) sqrt(tan(x)) + 1) - log(tan(x) + sqrt(2) sqrt(tan(x)) + 1))/(2 sqrt(2)) + constant
6a 1/2 (1 + sqrt(5))
+129845 6. The Golden Ratio is irrational.....so it won"t have an exact value
It equals ≈ 1.618 = "Phi" = [ 1 + sqrt(5) ] / 2
We can find an equation by using similar triangles in a regular pentagon
We have that
x^2 - x - 1 = 0 solving this with the quadratic formula gives
[ 1 ± sqrt (5) ] / 2
BTW : Interestingly, as the Fibonacci Series grows larger and larger, the ratio of a successive term to a preceding term approaches Phi
+129845 1 )
log3 ( 81 * (1/3)1/3 * (3 + ln 1)1/6 ) =
log3 ( 81 * (1/3)1/3 * (3 + 0 )1/6 ) =
log3 ( 81 * (1/3)1/3 * (3 )1/6 ) =
log3 ( 34 * (3)-1/3 * (3 )1/6 ) =
log3 (34 -1/3 + 1/6 ) =
log3 323/6 =
(23/6) log3 3 =
(23/6) * 1 =
23/6
+9673 Unsolved Problems:
3, 4(b), 7(why no one answer the easiest question), 8.
+129845 4b
y = x*sqrt (x) = x^(3/2)
Therefore....
y^(2/3) = x
So
x^2 + x + 1 =
[y^(2/3)]^2 + y^(2/3) + 1 =
y^(4/3) + y^(2/3) + 1
+129845 7
3xy = a + 2
xy = [a + 2 ] / 3
So
27xy =
27 [a + 2] / 3 =
[ 271/3 ] a + 2 =
[3 ] a + 2 =
3a * 32 =
9 * 3a
+129845 3
(a) I do not know the proof, but there is a trig identity that says that
a sin x + b cos x = c sin (x + α) where c = sqrt (a^2 + b^2) and
α = atan2(b,a)
So
c = sqrt (4^2 + 3^2) = sqrt (25) = 5 and
atan2 (b , a) = atan2( 3 , 4) ≈ .6435 (in rads)
4 sin x + 3 cos x = 5 sin ( x + .6435 )
(b) 4sin x + 3 cos x = 2
4sin x = 2 - 3 cos x square both sides
16sin^2 x = 4 - 12cosx + 9cos^2 x
16(1 - cos^2 x) = 4 - 12cos x + 9cos^2 x
16 - 16 cos^2 x = 4 - 12cos x + 9cos^2 x
25 cos^2x - 12 cos x - 12 = 0 let cos x = a
25a^2 - 12 a - 12 = 0
Using the quad formula.......
a = [ 6 ± 4 sqrt (21)] / 25
So
a = cos x = [ 6 ± 4 sqrt (21)] / 25
Wnen cos x = [ 6 + 4 sqrt (21)] / 25
arccos [ [ 6 + 4 sqrt (21)] / 25 ] = x ≈ .232 rads and ≈ 6.051 rads
When cos x = arccos [ [ 6 - 4 sqrt (21)] / 25 ]
arccos [ [ 6 - 4 sqrt (21)] / 25 ] = x ≈ 2.0866 rads and ≈ 5.228 rads
However.......232 rads and 5.228 rads are extraneous [ because of squaring ]
As the graph here shows.....the two solutions are x ≈ 2.0866 rads and x ≈ 6.051 rads
https://www.desmos.com/calculator/pjvewx15zw
