Since 5i=5eiπ2 And 2+i=5eix1 where x1=arctan⁡12, we have 5i2+i=5[cos⁡(π2−x1)+isin⁡(π2−x1)] Now from here how do I continue?

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Vasquez · Accepted Answer

We already have LHS=5ei(π2−x1),x1=arctan⁡12 (1) Continue with RHS=1+2i=5eix2, x2=arctan⁡2 (2) x1,x2 are complementary angles, x2=π2−x1→LHS=RHS Note: polar form represented by exponential form for simplicity in (1)(2). There is no difference between two forms since r and θ are same for LHS and RHS.

user_27qwe · Accepted Answer

Hint cos⁡(π2−x)=sin⁡x sin⁡(π2−x)=cos⁡x tan⁡(π2−x)=1tan⁡x So now let&#039;s develop the hint from where you left it i.e 5i1+2i=5(cos⁡(π2−x)+isin⁡(π2−x))=a+ib So a2+b2=5 and ba=tan⁡(π2−x)=1tan⁡x=2 And so 5a2=5 and a=±1, b=±2 The negative solutions are to be excluded because 0&amp;lt;π2−x&amp;lt;π2

nick1337 · Accepted Answer

Using polar form (with exponential notation for convenience), 5i2+i=5etπ25etarctan⁡12=5ei(π2−arctan⁡12)=5cos⁡(π5−arctan⁡12)+i5sin⁡(π2−arctan⁡12) Sketch the 1,2,5 right triangle and determine that the cosine of the relevant angle (which is complementary to the angle with tangent 12) is 15 and its sine is 25. The result above immediately simplifies to 1+2i.

nick1337 · Accepted Answer

cos⁡(5x)cos⁡(3x)−sin⁡(3x)sin⁡(x)=cos⁡(2x)12(cos⁡(2x)+cos⁡(8x))−12(cos⁡(2x)−cos⁡(4x))=cos⁡(2x)12(cos⁡(4x)+cos⁡(8x))−cos⁡(2x)=0cos⁡(8x)+cos⁡(4x)−2cos⁡(2x)=02cos⁡(2x)cos⁡(6x)−2cos⁡(2x)=0cos⁡(2x)[cos⁡(6x)−1]=0cos⁡(2x)=0→x=±π4+kπ discarded because this solution will make zero the denominator of the original equation cos⁡(6x)=1→x=kπ3

Since 5i=5e^{i \frac{\pi}{2}} And 2+i=\sqrt5e^{ix_1} where x_1=\arctan \frac12, we have

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