1 - Cos^2 X - Cos ^2 X = 1 - 2 C0s ^2X = 1 - 2 Cos^2 X . Upvote • 0 Downvote Add comment More. Report Still looking for help? Get the right answer, fast. Trigonometry. Solve for x cos (2x)=-1. cos (2x) = −1 cos ( 2 x) = - 1. Take the inverse cosine of both sides of the equation to extract x x from inside the cosine. 2x = arccos(−1) 2 x = arccos ( - 1) Simplify the right side. Tap for more steps 2x = π 2 x = π. Divide each term in 2x = π 2 x = π by 2 2 and simplify. To understand the cos2x formula, given solved examples show how cos 2x formula can be used. Example 1: Find the triple angle identity of the cosine function, using cos2x formula. Solution: cosine function’s triple angle identity is cos 3x = 4 cos3x – 3 cos x. cos 3x = cos (2x + x) = cos2x cos x – sin 2x sin x. Q 5. ∫ sin2x 1+cos2xdx. View Solution. Click here:point_up_2:to get an answer to your question :writing_hand:int cos2x dx. I having trouble solving this algebraically: Solve on the interval $[0,2\\pi]$: $\\cos(2x)+\\cos(4x)=0$. My problem is that I keep ending up with $3$ solutions: $\\pi/2, 3\\pi/2$ and $\\pi/6$. But whe lim x→0 cos(2x) 1 x lim x → 0 cos ( 2 x) 1 x. Use the properties of logarithms to simplify the limit. Tap for more steps lim x→0e1 xln(cos(2x)) lim x → 0 e 1 x ln ( cos ( 2 x)) Evaluate the limit. Tap for more steps elim x→0 ln(cos(2x)) x e lim x → 0 ln ( cos ( 2 x)) x. Apply L'Hospital's rule. Prove (1+sinx)(1-sinx)=cos^{2}x. en. Related Symbolab blog posts. My Notebook, the Symbolab way. Math notebooks have been around for hundreds of years. You write down Use the identity \cos^2(x)=\frac{1+\cos(2x)}{2}. You should then be able to solve this for x by way of the inverse. Floor value of infinite limit with sum. Trigonometry. Solve for ? square root of 2cos (2x)=1. √2cos(2x) = 1. Divide each term in √2cos(2x) = 1 by √2 and simplify. Tap for more steps cos(2x) = √2 2. Take the inverse cosine of both sides of the equation to extract x from inside the cosine. 2x = arccos(√2 2) Simplify the right side. Answer and Explanation: 1. The integral of 1 c o s 2 ( x) is tan ( x) + C, where C is any constant. To find this integral, we will make use of our definition of integrals along with the following two facts. 1 c o s 2 ( x) = s e c 2 ( x) sec 2 ( x) is the derivative of tan ( x ). First, we rewrite the integral using the first rule above. abIc6. You are using an out of date browser. It may not display this or other websites should upgrade or use an alternative browser. Forums Homework Help Precalculus Mathematics Homework Help What does (cos(2x))^2 equal? Thread starter justine411 Start date Apr 11, 2007 Apr 11, 2007 #1 Homework Statement (cos2x)^2Homework EquationsThe Attempt at a Solution I'm not sure if it is cos^2(2x) or cos^2(4x) or what. Should I use an identity to simplify it to make it easier to solve? Please help! :) Answers and Replies Apr 11, 2007 #2 What is there to solve??? (cos2x)^2 is just an expression. Apr 11, 2007 #3 In what sense is (cos(2x))2 a "problem"? What do you want to do with it? I will say that (cos(2x))2 means: First calculate 2x, then find cosine of that and finally square that result. Notice that it is still 2x, not 4x. The fact that 2 is outside the parentheses means that it only applies to the final result. Apr 12, 2007 #4 In what sense is (cos(2x))2 a "problem"? What do you want to do with it? I will say that (cos(2x))2 means: First calculate 2x, then find cosine of that and finally square that result. Notice that it is still 2x, not 4x. The fact that 2 is outside the parentheses means that it only applies to the final result. Doesn't (cos(2x))2 = cos2(2x)2 = cos2(4x2) ? Apr 12, 2007 #5 Doesn't (cos(2x))2 = cos2(2x)2 = cos2(4x2) ? No. 'Cos' is a particular operation and 2x is the argument. The exponent of 2 operates on cos, not on the argument. cos2y = cos y * cos y. There are also particular trigonometric identites with which one should be familiar, cos (x+y) and sin (x+y). Apr 12, 2007 #6 You still haven't told us what the problem was! Was it to write (cos(2x))^2 in terms of sin(x) and cos(x)? I would simply be inclined to write (cos(2x))^2 as cos^2(2x). Suggested for: What does (cos(2x))^2 equal? Last Post Jan 25, 2012 Last Post Nov 29, 2007 Last Post Jun 21, 2015 Last Post Apr 29, 2018 Last Post Sep 23, 2007 Last Post Apr 9, 2015 Last Post Feb 3, 2011 Last Post Sep 17, 2011 Last Post Apr 11, 2014 Last Post Jan 20, 2006 Forums Homework Help Precalculus Mathematics Homework Help Join Physics Forums Today! The friendliest, high quality science and math community on the planet! Register for no ads! \bold{\mathrm{Basic}} \bold{\alpha\beta\gamma} \bold{\mathrm{AB\Gamma}} \bold{\sin\cos} \bold{\ge\div\rightarrow} \bold{\overline{x}\space\mathbb{C}\forall} \bold{\sum\space\int\space\product} \bold{\begin{pmatrix}\square&\square\\\square&\square\end{pmatrix}} \bold{H_{2}O} \square^{2} x^{\square} \sqrt{\square} \nthroot[\msquare]{\square} \frac{\msquare}{\msquare} \log_{\msquare} \pi \theta \infty \int \frac{d}{dx} \ge \le \cdot \div x^{\circ} (\square) |\square| (f\:\circ\:g) f(x) \ln e^{\square} \left(\square\right)^{'} \frac{\partial}{\partial x} \int_{\msquare}^{\msquare} \lim \sum \sin \cos \tan \cot \csc \sec \alpha \beta \gamma \delta \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega A B \Gamma \Delta E Z H \Theta K \Lambda M N \Xi \Pi P \Sigma T \Upsilon \Phi X \Psi \Omega \sin \cos \tan \cot \sec \csc \sinh \cosh \tanh \coth \sech \arcsin \arccos \arctan \arccot \arcsec \arccsc \arcsinh \arccosh \arctanh \arccoth \arcsech \begin{cases}\square\\\square\end{cases} \begin{cases}\square\\\square\\\square\end{cases} = \ne \div \cdot \times \le \ge (\square) [\square] ▭\:\longdivision{▭} \times \twostack{▭}{▭} + \twostack{▭}{▭} - \twostack{▭}{▭} \square! x^{\circ} \rightarrow \lfloor\square\rfloor \lceil\square\rceil \overline{\square} \vec{\square} \in \forall \notin \exist \mathbb{R} \mathbb{C} \mathbb{N} \mathbb{Z} \emptyset \vee \wedge \neg \oplus \cap \cup \square^{c} \subset \subsete \superset \supersete \int \int\int \int\int\int \int_{\square}^{\square} \int_{\square}^{\square}\int_{\square}^{\square} \int_{\square}^{\square}\int_{\square}^{\square}\int_{\square}^{\square} \sum \prod \lim \lim _{x\to \infty } \lim _{x\to 0+} \lim _{x\to 0-} \frac{d}{dx} \frac{d^2}{dx^2} \left(\square\right)^{'} \left(\square\right)^{''} \frac{\partial}{\partial x} (2\times2) (2\times3) (3\times3) (3\times2) (4\times2) (4\times3) (4\times4) (3\times4) (2\times4) (5\times5) (1\times2) (1\times3) (1\times4) (1\times5) (1\times6) (2\times1) (3\times1) (4\times1) (5\times1) (6\times1) (7\times1) \mathrm{Radians} \mathrm{Degrees} \square! ( ) % \mathrm{clear} \arcsin \sin \sqrt{\square} 7 8 9 \div \arccos \cos \ln 4 5 6 \times \arctan \tan \log 1 2 3 - \pi e x^{\square} 0 . \bold{=} + Related » Graph » Number Line » Similar » Examples » Our online expert tutors can answer this problem Get step-by-step solutions from expert tutors as fast as 15-30 minutes. Your first 5 questions are on us! You are being redirected to Course Hero I want to submit the same problem to Course Hero Correct Answer :) Let's Try Again :( Try to further simplify Number Line Graph Hide Plot » Sorry, your browser does not support this application Examples \sin (x)+\sin (\frac{x}{2})=0,\:0\le \:x\le \:2\pi \cos (x)-\sin (x)=0 \sin (4\theta)-\frac{\sqrt{3}}{2}=0,\:\forall 0\le\theta<2\pi 2\sin ^2(x)+3=7\sin (x),\:x\in[0,\:2\pi ] 3\tan ^3(A)-\tan (A)=0,\:A\in \:[0,\:360] 2\cos ^2(x)-\sqrt{3}\cos (x)=0,\:0^{\circ \:}\lt x\lt 360^{\circ \:} trigonometric-equation-calculator cos^{2}x+2cosx+1=0 en $\begingroup$ Why: $$\cos ^2(2x) = \frac{1}{2}(1+\cos (4x))$$ I don't understand this, how I must to multiply two trigonometric functions? Thanks a lot. asked Oct 28, 2012 at 1:54 $\endgroup$ 2 $\begingroup$ Recall the formula $$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ This gives us $$\cos^2(\theta) = \dfrac{1+\cos(2 \theta)}{2}$$ Plug in $\theta = 2x$, to get what you want. EDIT The identity $$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ can be derived from $$\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B)$$ Setting $A = B = \theta$, we get that $$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \cos^2(\theta) - (1-\cos^2(\theta)) = 2 \cos^2(\theta) - 1$$ answered Oct 28, 2012 at 1:56 $\endgroup$ 1 $\begingroup$It’s just the double-angle formula for the cosine: for any angle $\alpha$, $\cos 2\alpha=\cos^2\alpha-\sin^2\alpha\;,$ and since $\sin^2\alpha=1-\cos^\alpha$, this can also be written $\cos2\alpha=2\cos^2\alpha-1$. Now let $\alpha=2x$: you get $\cos4x=2\cos^22x-1$, so $\cos^22x=\frac12(\cos4x+1)$. answered Oct 28, 2012 at 1:57 Brian M. ScottBrian M. Scott590k52 gold badges711 silver badges1179 bronze badges $\endgroup$ 1 $\begingroup$$$\cos(4x) = \cos^2 (2x) - \sin^2 (2x) = 2\cos^2 (2x) - 1$$ answered Oct 28, 2012 at 1:57 InquestInquest6,4472 gold badges32 silver badges56 bronze badges $\endgroup$ 0 Not the answer you're looking for? 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